'. II )
RKEIIY
*. .;y Of
UlfOKNIA
NEW MATHEMATICAL WORKS.
An Elemeiitaiy Treatise on the Differential otid Integral Calculus.
]{y (J. W. 1[i;mmin(;, M.A., Ffllow of St. John's College, C'anihiidpc,
bvo. bds. 9s.
Elementaiy Mechanics, accompanied by numerous Examples solved
Geometrically. iJy J. 15. Tnu-ui, M.A., Fellow and Mathematical Lec-
turer of Clare Hail. 8vo. bdt. 10*. 6rf.
A Short and Easy Course of Algebra.
Cliicfly designed for tlie use of tlic Junior Classes in Schools, -with a
numerous collection of Original Easy Exercises. By the Rev. T. Lund,
H.D., late Fellow of St. .John's College. 12mo. bds. 3a. 6d.
" His definitions are admirable for their simplicity and clearness."
Athenjeum.
" In order to ascertain how far the Author's performance comes nj) to fiis
design, ire /tate paid particular attention to those places where the learner
is most likely to stumble xtpon acknowledged difficulties. , . , In all
these toe have much reason to admire the happy art of the Azithor in
making crooked things straight, and rough places sfnoofh. The Student
must be hopelessly obtuse rcho does not, in folloxcing the guidance of
Mr. Lund, obtain increasing light and satisfactimi in every step of his
way; and such, too, is the strictly scientific as well as simple nature
of the course pursued, that he who makes hitnself master of it, will
have laid a firm foundation for an extensive and lofty superstructure
of mathematical acquirement,", — The Educator.
An Elementaiy Treatise on the Differential Calculus.
15y I. TuDiiuNTEu, M.A., Fellow of St. John's College. About Christmas .
This Work is intended for the use of Schools as well as for Students in
the Universities.
Plane Astronomy.
Including Explanations of Celestial Phenomena and Descriptions of
Astronomical Instruments. By the Kcv. A. R. Grant, M.A., FelloAv
of Trinity College. 8vo. bds. 6s.
The Principles of the Solutions of the Senate -House " Riders"
Exemplified in the Solution of those proposed in the years 1848 and 1851,
By F. J. Jameson, B.A., Caius College. In October,
A Treatise on Dynamics.
By W. P. Wilson, M.A., Fellow of St. John's College, and Professor
of Mathematics iji Queen's College, Belfast. 8vo. bds, 9s. Gd,
Cambridge: 3IACMILLAN AND CO.
London: GEORGE' BELL.
SOLUTIONS OF THE CAMBRIDGE
SENATE-HOUSE PROBLEMS
FOR FOUR YEARS :-1848-51.
N. M. FERRERS, B.A.,
OF GOXVILLE AND CAIUS COLLEGE, CAMBRIDGE,
AND
J. STUART JACKSON, B.A.,
ALSO OF GONVILLE AND CAIUS COLLEGE,
arambrtbge: MACMILLAN & Co.;
lEontJon: GEOKGE BELL; ©ubiin: HODGES & SMITH;
^(nbnrg^ : EDMOXSTON & DOUGLAS ;
(Glasgoto : J. MACLEHOSE.
1851.
LOAN STACK
CAMBRIDGE:
PHINTKD I!Y MF.TCAI.FE AND PALMl.I'.
PREFACE.
It will, we believe, be universally admitted that there is
no easier means of becoming acquainted with any branch of
Mathematics, than the study of Examples illustrative of its
principles. It is also indispensably necessary that the ingenuity
of the Student be thoroughly exercised in attempting to dis-
cover for himself the solution of any problem which may be
put before him : it is by no means our object, in publishing
this book, to save him the trouble of doing so. But we believe
that if, after having done his best to master a problem for
himself, he is still unsuccessful, he will then derive great benefit
from referring to the solution obtained by another person. It
is for this pui'pose that we hope the present collection will be
foimd of service.
Of the intrinsic value of these Problems we could have no
doubt, even if we knew less of them, coming as they do from
such high authorities. We have as little doubt that they are
on all accounts the best problems that we could have chosen
for solution for the benefit of the Student, for their general
value, their variety, and because they shew what Senate-
house Problems are, and arc likely to be in coming years.
418
IV PREFACE.
Part 1. contains the solutions of those proposed in tlic first
three days of examination : they are of a simpler kind than
those proposed in the remaining five days, the solutions of which
form Part II. llie solutions of many problems have been
kindly timiished by the Moderators by whom they were pro-
posed: we take this opportunity of returning om* acknowledg-
ments to them and others of our friends who have assisted
us in the progress of the work. We shall also feel much
obliged to any of our readers who will send us either correc-
tions of our solutions or improvements upon them.
Some difficulty, it will easily be understood, has been found
in bringing all the problems to appear m their right places:
any problems, however, which have been omitted in the body
of the work, will be found in the Appendix, The problems
on pp. 226, 241, should have been placed among the Trigo-
nometry of 1850, and the Geometry of Three Dimensions of
1851, respectively.
C O IS T E N T S.
PART I.
Euclid .
Algebra
Trigonometry
Conic Sections
Statics .
Dynamics
Newton
Hydrostatics
Optics .
Astronomv •
Page
1
9
19
25
32
37
48
53
59
64
PART II.
Euclid .
Algebra
Plane Trigonometry
Spherical Trigonometry
Theory of Equations
Geometry of Two Dimensions
Differential Calculus
Integral Calculus
69
74
98
107
112
110
171
183
i CONTUNTS.
Page
Geometry oH Three Dimensions . . . 1*«^3
Ditferential Equations .... 222
Definite Integrals .... 233
Calculus of Finite Differences .... 240
Statics ..... 245
Dynamics of a Particle .... 2*37
Rigid Dynamics . . . . • 283
Hydrostatics ..... 323
Hydrodynamics ..... 333
Geometrical Optics ..... 339
Astronomy . . • • • 3J5
Disturbed Motion ..... 362
Attractions ..... 36j
Physical Ojjtics ..... 375
Calculus of Variations .... 380
Appendix ...... 384
PART I.
SOLUTIONS OF THE SENATE-HOUSE PROBLEMS.
EUCLID.
ERRATA.
I'AGK LINE /. ; J ,.'
87, 5 from bottom, for p read n .
95, 10, for (i,,.,n"p ^''^"'■^ "7'---i + "^•
— 17, for ... t <v.,)-i Vi^ '■'"'^^ •■•■^ "p 0 Vi-
176, 12, /or /t' - 2rt/ rcat^ A' - 2a A.
288, 8,"/o?- -R' J-ertc/ By.
215, 1, supply reference to figure (S8).
23l', !,>'• (tig- 08) 7-m(/ (tig. 89).
Figure 4f5, /b;- j> 2/ s »e'^t^ ~ i^ y-
Again, because EF is a diameter of the circle, therefore the
angle FCE is a right angle. But CE bisects the inght angle
ACB, therefore ACE is half a right angle, therefore also FCA
is half a right angle ; that is, FC bisects the supplement of the
right angle ACB.
2. A, B, C, (fig. 2) are three given points in the circum-
ference of a circle ; find a point P, such that if AP, BP, CP
meet the circumference in D, E, F, the arcs DE, EF, may be
equal to given arcs.
Join AB, and on it describe a segment of a circle, containing
an angle equal to the sum of those subtended at the circiim-
SOLUTIONS OF THE SENATE-HOUSE PROBLEMS.
EUCLTD.
1848.
1. If the hypotlienuse AB (fig. 1) of a right-angled triangle
ABC be bisected in D, and EDF drawn pei-pendicular to AB,
and DE, DF cut off each equal to DA, and CE, CF joined ;
prove that the last two lines will bisect the angle at C and its
supplement respectively.
Join CD, then shall CD be equal to half the hvpothenuse
AB, that is, to DE or DF ; therefore a circle described from
centre D, with radius DC, will pass through B, E, A, F. Let
this circle be described, then the angle ECB, at the circum-
ference, is equal to half the angle EDB at the centre, that is
to half a right angle, and therefore to half the angle ACB ;
that is, CE bisects the angle ACB.
Again, because EF is a diameter of the circle, therefore the
angle FCE is a right angle. But CE bisects the right angle
ACB, therefore ACE is half a right angle, therefore also FCA
is half a right angle 5 that is, FC bisects the supplement of the
right angle ACB.
2. A, B, C, (fig. 2) are three given points in the circum-
ference of a circle ; find a point P, such that if AP, BP, CP
meet the circmnference in D, E, F, the arcs DE, EF, may be
equal to given arcs.
Join AB, and on it describe a segment of a circle, containing
an angle equal to the sum of those subtended at the circum-
H
2 SOLUTION OF SENATE-HOUSE PHOBLEMS. [1849.
forence by the arcs AB and DE. Also join BC, and on it
describe a segment of a circle containing an angle equal to the
sum of those subtended at the circumference by the arcs BC
and EF. These segments shall intersect in the required point P.
For join AP, BP, CP, and produce them to meet the cir-
cumference in D, E, F, respectively. Join AE, then the angle
EAD is equal to the difference of the angles APB, AEP,
that is, to the angle required to be subtended by the arc DE.
Therefore DE is equal to the arc required. Similarly it may be
shewn that EF is equal to the arc required, and therefore P is
the required point.
1849.
1. Through a point C (fig. 3) in the eirciunference of a circle,
two straight lines ACB, DCE, are drawn, cutting the circle in
B and E ; prove that the straight line which bisects the angles
ACE, DCB, meets the circle in a point equidistant from B
and E.
Let CP be the line bisecting the angles ACE, BCD ; P the
point in which it meets the circle. Join PB, PE, BE ; then
because the angles PBE, PCE are in the same segment, there-
fore they are equal to one another.
Again, because the angles BCP, BEP are opposite angles
of a quadrilateral inscribed in a circle, therefore they are
together equal to two right angles, that is to ACP and BCP.
Therefore, taking away the common angle BCP, ACP is equal
to BEP. But ACP is equal to ECP by constniction, therefore
from above ACP is equal to EBP : and it has been shewn to be
equal to BEP, therefore the angles EBP, BEP are equal to
one another, therefore PE is equal to PB. That is, the point P
in which the bisecting line CP meets the circle is equidistant
from B and E.
2. Two circles intersect in A and B (fig. 4). At A, the
tangents AC, AD are drawn to each circle and tenninated by
the circmnference of the other. If BC, BD be joined, shew
that AB, or AB produced if necessary, bisects the angle CBD.
1849.] EUCLID. 3
Produce CA, DA, to E, F. Then the angle CAF is equal
to the angle DAE : but the angle CAF is equal to the angle
ABC in the alternate segment, also the angle DAE is equal to
the angle ABD in the alternate segment. Therefore the angles
ABC, ABD are equal to one another, and AB, produced if
necessary, bisects the angle CBD.
3. Draw a line to touch one given circle, so that the part
of it contained by another given circle may be equal to a giveii
straight line, not greater than the diameter of this latter circle.
Let ABC, DEF (fig. 5) be two given circles ; it is required
to draw a straight line touching the circle ABC, so that the
part of it contained by DEF may be equal to a given straight
line, not greater than the diameter of DEF.
In the circle DEF place the straight line DE, equal to the
given straight line. Find G the centre of this circle, and with
G as centre describe a circle touching DE. Draw AFH a
common tangent to this latter circle and ABC, cutting DEF
in F, H, this shall be the line required.
For since FH, DE each touch a circle whose centre is G,
therefore they are equidistant from G, the centre of the circle
DEF. Therefore FH is equal to DE.
Hence AFH is drawn touching the cii-cle ABC, and the
part of it contained by DEF is equal to the given straight line.
4. A quadrilateral figure possesses the following property :
any point being taken, and four triangles fonned by joining this
point with the angular points of the figure, the centres of
gravity of these triangles lie in the circumference of a circle ;
prove that the diagonals of this quadrilateral arc at right angles
to each other.
Let ABCD (fig. 6) be the quadrilateral, P any point within
it : E, F, H, K, the middle points of the sides. Join PE, PF,
PH, PK. And in them take PG„ PG,, PG„ PG,, respectively
equal to two-thirds of PE, PF, PH, PK. Gfi.jGjOc^ shall be
the centres of gravity of the triangles PAB, PBC, PCD, PDA.
Join EF, FH, HK, KE, these lines are evidently parallel to
b2
4 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
^^.^\i C^^Gs, G3G,, G,G, ; and therefore E, F, H, K, lie in the
cireumference of the same circle. But EF, HK are each
parallel to the diagonal AC, therefore also to each other.
Similarly FH, EK arc parallel to each other, therefore EFHK
is a parallelogram. And since it is inscribable in a circle, each
of its angles is a right angle. Therefore also the diagonals
AC, BD, which are respectively parallel to the sides of the
parallelogram, are at right angles to each other.
1850.
1. If ABCD (fig. 7) be a parallelogram, and P, Q two
points in a line parallel to AB, and if PA, QB meet in E,
and PD, QC in S, prove that RS is parallel to AD.
Because CD is parallel to QP, therefore SD is to SP as CD
to PQ. And because AB is parallel to PQ, therefore RA is to
RP as AB to PQ. But AB is equal to CD, therefore RA is to
RP as SD to SP, therefore RP is to AP as SP to DP, there-
fore RS is parallel to AD.
2. Two sides of a triangle, whose perimeter is constant, are
given in position ; shew that the third side always touches a
certain circle.
Let ABC (fig. 8) represent the triangle ; AB, AC being the
sides given in position. Describe a circle DEF touching BC
and AB, AC produced. The side BC shall always touch the
circle DEF.
For since BD, BF both touch the same circle, therefore BD
is equal to BF. Hence AD is equal to AB, BF together.
Similarly AE is equal to AC, CF together.
Therefore AD, AE together are equal to AB, AC, BC
together ; that is, to the perimeter of the triangle ABC, which
is constant. But since AD, AE touch the same circle, therefore
AD is equal to AE, and their sum has been shewn to be con-
stant; therefore AD, AE are each constant, that is, the circle
touching BC, and AB, AC produced, touches AB, AC in fixed
points ; that is, it is a fixed circle. Therefore BC always
touches a fixed circle.
1851.] EUCLID. 5
1851.
1. In AB, the diameter of a circle, take two points C, D,
equally distant from the centre, and from any point E in the
circmnference draw EC, ED ; shew that
EC' + ED'' = AC + AD^
Take O (fig. 9) the centre of the circle, and join EO, and
di'aw EF pei'pendicular to AB.
Then because CD is bisected in O, and produced to A,
.-. AC' + AD' = 2 (AC + OC) [Em. ii. 10).
Again, because EC is opposite to an acute angle O of a triangle
ECO, therefore
EC + 20C.0F = EO' + OC [Euc. ii. 13).
And because ED is opposite to the obtuse angle O of a triangle
EOD, therefore
ED'' = EO' + OD' + 20D.0F [Euc. ii. 12),
= EO' + OC + 20C.0F ;
.-. EC^ + ED' = 2 (EC + OC),
= 2(A0' + 0C),
= AC 4- AD' from above.
2. If through the fixed points P, Q, (fig. 10) parallel lines be
drawn meetmg two fixed parallel lines in the points M, N ;
then the line through the points M, N, passes through a fixed
point.
Join PQ, and let it meet MN in O, and the given pair of
parallels in Ii, S, O shall be a fixed point.
For since QN is parallel to PM, therefore QO is to PQ as
ON to NM. And since NR is parallel to MS, therefore OR is
to RS as ON to NM. Therefore OR is to RS as OQ to QP,
or OR is to OQ as RS to QP ; that is, QR is divided in a con-
stant ratio in O, and therefore O is a fixed point.
3. In a given circle it is required to inscribe a triangle,
similar and similarly situated to a given triangle.
6 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Let ABC (rig. 11) be a given triangle, DEF a given circle;
it is required to inscribe in DEF a triangle, similar and similarly
situated to ABC.
At the point A in the straight line AB, make the angle
BAG equal to the angle ACB. Find H, the centre of the
circle DEF, and draw UK parallel to AG, HD pei*pcndicular
to HK. Through D draw DE, DF parallel respectively to
AB, AC, DEF shall be the triangle required.
For di'aw DL parallel to HK or AG, and therefore touching
the circle at D. Then the angle LDE is equal to the angle
GAB. But GAB is by constniction equal to ACB, and LDE
is equal to DFE in the alternate segment. Therefore the angle
DFE is equal to ACB. Similarly the angle DEF is equal to
the angle ABC. Therefore the remaining angle EDF is equal
to BAC, so that the triangles ABC, DEF are similar. And
since the sides DE, DF are parallel respectively to AB, AC,
therefore EF is parallel to BC, and they are similarly situated.
4. Describe a circle which shall pass through two points,
and cut off from a given straight line a chord of given
length.
Let A, B (fig. 12) be two given points, CX a given line,
it is required to describe a circle passing tlu'ough A, B, and
cutting off from CX a chord of given length.
Join BA, and produce it to meet CX in C. Bisect AB in E,
and draw EF perpendicular to AB. With A as centre, and
radius equal to half the required chord, describe a circle FGH,
cutting EF in F. With F as centre, and FA as radius, describe
a circle BAG. Join CF, and let it cut the circle BAG in K, L.
From CX cut off CM, equal to CK. The circle described
through A, B, ]\I shall be the circle required ; that is, if N be
the second point in which it meets CX, MN shall be equal to
the required chord.
For the rectangle CM.CN is equal to CA.CB by the pro-
perty of the circle ABM. And the rectangle CA.CB Is equal
to CK.CL by the property of the circle KAB. Therefore the
rectangle CK.CL is equal to CM.CN.
1851.] EUCLID. 7
But CK is equal to CM, therefore CL is equal to CN, and
therefore KL is equal to MN, and KL is equal to the required
chord, therefore MN is so, and the circle ABNM is the circle
required.
5. Give a constniction* for finding the common tangents of
two circles, and shew that if through the intersection O of two
of the common tangents which meet in the line joining the
centres of the two circles, there be drawn a transversal meeting
the circles in A, A', and B, B', respectively, then (the points
denoted by B, B' being properly chosen) OA.OB' = OA'.OB is
independent of the position of the transversal.
Let ABC, DEF (fig. 13) be two circles, whereof DEF is the
greater; find G, H, their centres, and with H as centre, and
radius equal to the differencef of the radii of the given circles,
describe a circle. Through G draw GK, a tangent to this
circle ; di-aw GA perpendicular to GK, and AD perpendicular
to GA, meeting DEF in D : AD shall be a common tangent.
Let C, D (fig. 14) be the points of contact of one of the
coimnon tangents. Then we easily see that
OA:OA' :: OB: OB',
.-. OA.OB' = OA'.OB ;
and also OA.OB' : OA.OA :: OB.OB' : OA'.OB,
or OA.OB' : OC^: OD^ : OA'.OB,
::0D^ OA.OB',
.-. OA.OB' = OA'.OB = OC.OD,
which is independent of the position of the transversal.
6. Shew that a triangle made to revolve in the same direc-
tion about its three angular points in a proper order through
angles double of the angles of the triangle at the same angular
points, will return to its original position.
* A demonstration of this construction is not required.
t We might also take a radius equal to the sum of the radii of the given
circles, in which case the common tangent would touch the two circles on
opposite sides of the line joining their centres.
8 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1851.
Let ABC (fig. 15) denote the triangle in its first position.
At the point A, in the straight line AB, make the angle BAG'
equal to the angle BAG, and make AG' equal to AG. Again,
make the angle GAB' equal to the angle BAG', and AB' equal
to AB. Join B'G' ; then in the triangles GAB, G'AB', the sides
GA, AB are respectively equal to G'A, AB', and the angle
GAB is equal to the angle G'AB' ; therefore the triangles are
equal in all respects. And the angle B'AB is double of the
angle GAB, therefore G'AB' is the position of the triangle after
revolving round A through an angle equal to 2. GAB.
Again, join BG' ; make the angle BG'A' equal to the angle
B'G'A or BGA, and G'A' equal to G'A. Join A'B ; then in the
triangles AG'B', A'G'B, the sides AG', G'B' are respectively
equal to A'G', G'B, and the angle AG'B' is equal to A'G'B.
Therefore the triangle A'BG' is equal in all respects to AB'G',
and therefore to ABG. And the angles BG'A', AG'B are
together double of B'G'A or BGA, therefore A'BG' is the posi-
tion of the triangle after revolving round the angles A, G.
Again, since the angles G'BA', G'BA, GBA, are all equal,
GBG' is double of A'BG' or ABG. Therefore the triangle, after
revolving romid its three angular points in succession, through
angles double of the angles of the triangle at those points,
returns to its initial position ABG.
( 9 )
ALGEBRA.
1848.
1. A ship sails with a supply of biscuit for 60 days, at a
daily allowance of 1 lb. a-head : after being at sea 20 days she
encounters a storm, in which 5 men are washed overboard, and
damage sustained that will cause a delay of 24 days, and it is
found that each man's allowance must be reduced to f lb. Find
the original number of the crew.
Let X = the original number of the crew.
Then 60a; = number of lbs. of biscuit with which they started.
4:0a; = remaining after being at sea
20 days.
And the remainder of the crew = a; — 5, who have to remain at
sea for 64 days, under a daily allowance of f lb per man.
.-. f 64 (a; - 5) = 40a;,
.-. 64a; - 320 = 56a;,
.-. 8a; = 320,
and X — 40,
the original number of the crew.
2. If a, i, and x be positive, and a > b, prove that
X + a X -\- b
> <
(a;" + a^)i {x' + h'
,,, , x -\- a X + b
W e have t-^. ^. > <
yjTi , according as a- > < [ab)
[x' + ay (a;* + b'J '
x^ + 2aa; + a" x^ + 2bx + h
as = :, — > <
X* + d' x^ + W '
2aa; ibx
x' + «' x' + /'' '
10 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
a h . . ..
as -5 r. > < -5 Ti , Since x is positive,
as [a — h) x'' > < d'h — a//,
as (a — h) x^ > < ah [a — Z>),
as x^ > < ai, since a is greater than b,
or as X > < («&)*.
3. If a, ^, c, be in haiinonic progression, and ti be a positive
integer, shew that a" + c" > 25".
Suppose a, 5, c, all positive.* Then we have
- , T 1 - •) in aritlinietic progression,
a b c
2ac
but a + c > 2\/«c,
2ac ^ ,
.*. -^— > 2v«c,
0
or V^c > 5,
but a" + c" > 2a*"c*",
a fortiori a' + c" > 2V.
1849.
1. Reduce to its simplest fonn the expression
(1 - gQ (1 - b'') (1 - <f) - (g + be) [b + cff) (c + a^)
1 - a' - ^*' - c' - 2abc
We have (l-a'') (1-Z*^) [l-c')
= 1 _ a^ _ ^,'^ _ c^ + J'-^c^ + c'-'a'-' + o;'// - d'bV
[a + be) [b + CO) (c + (ib)
= aba + ¥c' + 6'ci' + c^W + [c^ + 5^ + e') abc + d'V'c^ ;
* If one of these three quantities were negative, the proposition enunciated
Avould not be necessarily true.
1850.] ALGEBRA. 11
... (1 _ a') (1 - h'') (1 - 6') -{a + he) [h + m) (c + ab)
= l-d'- ¥ - c' - abc - {a' + 1/ + 6') abc - 2dVc'
= [\-ci'- h' - c' - 2abc) (1 + abc),
(1 _ d^) (1 - i;^) (1 _ c') -{a + be) [b + m) (c + ah)
' ' i-ce-h''- & - 2abc ~ ^ + ''^''•
2. Find a whole number which is greater than three times
tlie integral part of its square root by miity. Shew that there
are two solutions of the problem, and no more.
Let X be the integral part of the square root,
x^ + y the whole number.
Then, by the conditions of the problem,
a;^ + y = 3a; + 1,
.•. a;'' — 3a; = 1 — ?/,
.-. x' _ 3aj + f = 1,3 _ y^
Now y is essentially positive, and in order that x may be
real it is necessary that y be less than 5'. Also, in order that x
may be an integer, i}^ — y)^ must be of the form |-(2?/i+l),
m being some integer, such that [2m + l)'"* is less than 13.
Therefore the only admissible values of m are 0 and 1. Hence
a; = i + ^ori + i,
= 2 or 3.
Therefore 3a; + 1 = 7 or 10,
and 7 and 10 are the only solutions of the problem.
1850.
1. A number of persons were engaged to do a piece of work
which would have occupied them m hours if they had com-
menced at the same time; but instead of doing so, they com-
menced at equal intervals, and then continued to work till the
whole was finished, the payment being proportional to the work
done by each ; the first comer received r times as much as the
last : find the time occupied.
12 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
Let X be the number of persons employed, y the nmnber of
hours the last worked, z the interval, in hours, between the first
and second person commencing work, which is also that between
the second and third, and so on. Then, by tlie conditions of the
problem,
y + [x-l) z = ry (1).
Also, if W be the work done in an hour,
[?/+ {y-\-z) + {y+^z) + ... + {?/+ [x—\) zW W= whole work done,
•*• y + (i/+^) + (y+2s) 4- ... + {y + (a;-l) s} = mx,
or [2y + (a? — 1) z] \x = mx^
.'. 2y -\- {x — l)z = 2m,
and, by (1), y + {x-l) z = ry,
.'. y = 2m — ry,
Therefore whole time occupied = time from the first person
begiiming to work till the completion of the work
= y + {x-l)z,
= ry by (1),
2mr , , . ,
= hours by 2).
r + 1 -^ ^ ^
2. Shew that the product of the terms of an arithmetical
progression is greater than [alf" ; and that the sum of the terms
of a geometrical progression is less than (a+?)^«; where in
both cases a, /, and n denote the first and last terms, and the
number of terms respectively.
(a) Let h be the common difference in the arithmetical
progression, P the product of its terms, then
P=a{a + b) {a + 2h) ... {a + {n-l)b},
and I = a + [n — 1) b.
Now (a + mb) [a -f- {n- )ti—l) b} = a^ + (n — 1) ab + (n—m — 1) mh^.
1851.] ALGEBRA. 13
This will be least when m = 0, negative values of m being
excluded; but in that ease
{a + mh) [a -\- (n — m — \) b] = al-
therefore the product of any two tenns equidistant from the
mean is not less than al.
The property enunciated follows at once from this when n is
even. If 7i is odd, the middle term is
71 — 1 -
and (a + ^^-^ h\ = «' + {n - 1) ah + \^-^ o\ ,
> d^ + [n— 1) «J,
> al ;
.'. a + ^-^ h > {aI)K
Hence, in all cases, P > [al)-".
(/3) Let r be the common ratio in the geometrical pro-
gression, S the sum of its tenns ; then
b = a — — ,
r — 1 '
and I = ar"'^^
.'. a + l = a{l+ r"-').
Now ar"^ + ar"-"-' - (a + 7) = « ('•'" - 1) - « (^''"' - »•""""'),
= « (r - 1) (1 - r ),
which is negative for all positive values of m.
Hence the smn of the first and last tenns is greater than the
sum of any other two tenns equidistant from the mean.
The property enunciated follows at once from this when n is
even. If n be odd, the middle tenn is
and ri^"-'> < ^^f^,
,,„_!) a + I
2
Hence, in all cases, S < (a + /) ^n.
14 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
1851.
3.r — 2
1. If , -xV 7:^r} ttn be expanded in a series ascend-
[x— 1) [x — 2) [x — 6) ^
lug by powers of a:, find the coefficient of x\
This is best effected by resolving the given expression mto
its partial fractions, and expanding each fraction separately.
For this piu'}30se, assume
3a; -2 A B G
{x-l) {x-2)[x-^)~ x-l'^ x-2'^ x-^"*
A, Bj C being independent of x.
ThenSx-2=A{x-2){x-3) + B{x-3){x-l) + C{x-l){x-2)
Hence, putting a; =1, identicalhj.
3.1 - 2 = ^(1-2) (1-3),
or 1 = 2A (1).
Similarly, putting a; = 2,
3.2 - 2 = 5(2-3) (2- 1),
or 4 = - B (2),
and putting a^ = 3,
3.3 - 2 = 0(3-1) (3-2),
or 7 =2C (3);
Sx-2 _ 1 4_ 7
•'' (a;-l)(a;-2)(a;-3) ~ 2 (a; - 1) a; - 2 "^ 2 (a; - 3) '
11 2 7 1
2"^ u J. — J
2 1-a;l-^a; 6 1-ia;'
= -i(l+a;+ ... +a;"+...)
+ 2{l+^x-}- ... + {^xY +
-l{l + ^x + ... + i^xr + ...],
in which the coefficient of a;"
_ _ 1 2 _ 7 1
~~2'^2^~6 3""'
_ J 1. _ 7 1
- 2«-i 2 2 3^"
the required coefficient.
1851.] ALGEBRA. 15
2. Find the siun of the different numbers which can be
formed with m digits a, n digits /3, &c., the entire series of
w + » + &c. digits being employed in the formation of each
number.
The total number of numbers which can thus be fonned, is
equal to the number of permutations of m -^-n -\- things,
whereof wi are of one kind, n of another..., taken all together,
_ 1.2...(m + /i + ...)
~ (1.2. ..w) (1.2...??)... '
Now, the number of times a will be found in any assigned
place : the nmnber of times j3 will be found there :...'.: m '. n : ... \
therefore the number of times a will be found there
m 1.2...(w + w + ...) 1.2...(7n + »i+...-l)
— = m
??i + « + ... [1.2... m) [1.2... n)... (1.2...w) (1.2...w)... '
Similar expressions holding for the number of times y3, 7,... ■will ■
be found there : we have, if S be the sum of the digits in any
assigned place,
[ma + n^+ ...) \.2...[m + n-\- ... - 1) _
[1.2... m) [1.2... n)... '
therefore if 2 denote the sum of all the numbers,
2 = ;S(1 + 10 + 10''' + ... + iO"'+"+--^),
= ^ 9 '
_ 10»'+"^--_l (»;a + 7?/3+...) 1.2...(7>? + »+•..-!)
~ 9 [1.2... m) [1.2... n)... '
the sum required.
3. The difference between the arithmetic and geometric
means of two niunbers is less than one-eighth of the squared
difference of the numbers divided by the less number, but
greater than one-eighth of such squared difference divided by
the greater nimaber. If a*, y be any two nimibers, a-^, y, their
arithmetic and geometric means, a:^, y^ the anthmetic and geo-
metric means of iCj, 3/,, and so on, find major and minor limits
for the difference ic„ - ?/„.
16 SOLUTIONS OF SENATE-HOUSE PKOliLEMS. [iHol.
(a) Taking the notation of the Litter part of the problem,
we have
x-2 (a-M)i + y (x* - iAY
• • ^, ^1 - 2 ~ 2 *
Now, one-eighth of the squared difference of the numbers
K-yJ
8
4
And [x^-\-y'^y lies between [2y^f and (2a;*)'\
or > 4?/, < 4a? ;
4
>
3/ 8 '
1 [x-yY
X 8
(/3) From above it appears that
„ _ „, ^ (^«-i ~ 3/«-i) ^ (^«-i ~ 3/»-i) .
therefore, a fortiori^
X — V < ^ (^"-2 ^/n-a)
2 '
< >
<
{^-yf
1
(^«-2 3/«-2)
8^«-,
. (8^.-2)^ '
(^-2/f
%„-.(83/„-2)^"(%)''"' 8a.,^_,(8.0^..(8a.)^"-^
4. If all the sums of two letters that can be formed with
any n letters be multiplied together, then in each term of the
product, the sum of any r of the indices cannot exceed the
number rn — \r[r-\-\).
1851.] ALGEBRA. 17
Let rtj, a.^, f/^^, be the letters.
Then the product will be of the form
Now, any one of the letters, as «|, only appears n—\ times in
this product, that is, once in each of the factors a^ + a.^ . . . a, + r/,^,
therefore in no term can its index be greater than n—\.
And in such terra, the index of a^ cannot be greater than
w— 2, for a,^ can only enter n—2 times as a factor of such a tenn,
the factor a, + a.^ being excluded.
Similarly, in the term involving
a"~^a^~'^ the index of a^ cannot be > 7? — 3,
>i-l n-2 n-r+l
ttj rtj ...(i,_^ a,. > ?« — r,
therefore the sum of any r of the indices cannot be greater than
(m-1) + (n-2) +... + [n-r),
= rn - -^r (/•+!),
the required limit, which the sum of the r indices cannot exceed.
5. Eliminate x from the equations
[x — a) [x — h) = [x — c) (x — d) = [x — e) [x —f)^
and from the same equations with the additional relation e =/",
find a quadratic equation for determining the quantity e or f.
Shew also that if ?«', vi" be the values of e or f^ then m" — m is
a harmonic mean between a — in\ b — m\ and between c — m\
d — m.
(a) Since [x — a] [x — h) = [x — c) [x — d) = [x — e) [x—f),
we get
x^ — [a + b) X + ah = x'^ — [c + d) x + cd = x^ — {e +f) x + ef]
.'. ie +f— a — b) X = ef — ab^
[e +f — c — d) X = ef — cd^
... [ef- cd) [e +f- a-b) = [ef- ab) {e +/'- r - d). ..{!),
18 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
an equation from which x is eliminated, and which may be put
in the more symmetrical form
ab{c + d-e-f)+ccl[e+f-a-h) + ef{a + h-c-(l)=0...{2).
(/9) If (' =/, equation (1) becomes
[e' - cd) {2e - a - b) = {e' -ah){2e-c- d),
... i^c-\-d-a-h) i + 2[ah - cd) e -\- {a + b) cd - [c -\- d) ab = 0.. .(3),
the quadratic for the deteniiination of e or/.
(7) If m'^ m" be the roots of equation (3),
ab — cd
m + m" = 2
mm
a + b — c — d''
ab {c + d) — cd [a + b) ^
a + b — c — d '
{ab — cd) {a + b)
.'. 2 (ah + m'm") =2 , , ,
^ ' a + b - c -d ^
= (a + b) [m + m").
Now
2 {ab->rm'm') — {a + b) {m' + m") = [b—m') {a—m") — {a — m') {m"— b),
whence {a — on) {m" — b) = {b — m) {a — m")^
.'. a — m' '. b — m! '.'. a — m" : m" — b^
or a — m' : b — m :: a — m — {m" — m') : m" — m — {b — m')^
whence m" — in' is a harmonic mean between a — m\ b — on.
Similarly, it is a harmonic mean between c — on, d — on.
19
TRIGONOMETRY.
1848.
1. The angles of a quadrilateral inscribed in a circle taken
in order, when multiplied by 1, 2, 2, 3, respectively, are in
Arithmetical Progression; find their values.
Let 6^ (j) be two adjacent angles of the quadrilateral, then
TT — 6, TT — (j), will be the angles respectively opposite to them ;
and, by the conditions of the problem, 0, 2^, 2(7r — 6), 3(7r — 0 1,
are in Arithmetical Progression.
.-. 2(f)- e = 2(7r-^) - 2(f) = S[7r-(f)) - 2(7r-^),
.-. A(f) + e = 27r,
4^ - (/) = TT,
.-. 17^= Gtt,
17<^ = Ttt,
17 ' ^ 17 '
a llTT , IOtt
17 ' ^ 17 '
the required values of the angles of the quadrilateral.
2. Prove that sin 3^ sin'^ + cos3^ cos'^ = cos^2^.
We have cos 3^ = 4 cos"^ — 3 cos ^,
sin 3^ = 3 sin^ - 4 sin'^;
.-. sin 3^ sin'^ + cos 3^ cos'^
= 3 (sin*^ - cos'<9) + 4 (cos«^ - sin"^),
= 3 (cos^^ + sin''^) (sin'(9 - cos*^) + 4 (cos«^ - sin"^),
= cos'^ - 3 cos'^ sin'-'^ + 3 cos'^ siu'(9 - sin"!?,
= (cos'^-sm''^)=',
= cos" 2^,
the required result.
C2
20 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1848.
3. Having given the three right lines di'awn from any point
to the three angular points of an equilateral triangle, determine
a side of the triangle.
Let ABC (fig. 16) be the triangle, 0 the point from which
the lines are dra^vn, OA = a, OB =5, 0 0 = c ; also let the
angle BA 0=0^ CA 0 = (f>^ and let a side of the triangle = x.
Then, by the triangle BA 0,
x' + d^ - 2ax cos6 = If (1).
By the triangle CA 0,
x^ + d^ — 2ax cos(f) = c^ (2).
Also 6 + (f) = ^TT.
Adding (1) and (2), we get, observing that cos ^ + cos ^
^ 0-\- cf> e-<f>
= 2 cos — — ^ cos —!- ,
2(a;^ + a') - ^ax cos^tt cos — -^ = W + c' (3).
Subtracting (2) from (1), and observing that cos ^ — cos ^
= 2 sm — - sm — ^ — - ,
Q JL
Aax sin^TT sin — — -^ = b^ — c^ (4).
By (3) and (4),
=^ ^^
COS^ ^TT sin""' ^TT
,,,'^ ji+i^\'^-:±^ ^ac^ff ,
... [y^ + c^ _ 2d'y - ix\W + 6' - 2d' + ?>a^) -h 3 (// - d')' + ^x' = 0,
... a;* _ [a' + P + c') x' + a' + b* + c*- {bV + c'd' + d'F) = 0,
2 d' + b^ + d' {{d'+b^+Cy , 4 , ,4 , 4N , 72 2, 2 2, 271!
.-. x'= ± < ^ — (a* -I- b* + c) + Fc+d'ar^-a'b'
= '"'"^t'"^''' ± % [2 (?>V + cV + d'h') - [a' + b' + c*)}4,
which determines a side of the triangle.
1849.] TRIGONOMETRY. 21
1849.
1. If 0 — a, 0, 0 + a, be three angles whose cosines are in
Harmonical Progression, prove that
cos (f> = 2^ cos ^a.
Since cos(^ — a), cos</), cos(^ + a) are in Harmonical Pro-
gression, we have
2 1 1
+
cos <f> cos {(}> + a) cos ((jb — a) '
2 cos<^ cos a
cos (0 — a) cos (^ + a) '
/. cos^<^ cosa = cos(^ — a) cos((^ + a),
= ^ (cos 2(f) + cos 2a),
= cos^^ — sin'^a,
2 , sin'^'a
.*. cos © = ,
1 — cosa
4 sin'"* ^a cos'"* ^a
~ 2 sin' ^a '
= 2 cos"''^a,
.*. cos<^ = 2- cos^a,
the required relation.
2. A person wishing to ascertain his distance from an in-
accessible object, finds three points in the horizontal plane at
which the angular elevation of the summit of the object is the
same. Shew how the distance may be found.
Let 0 (fig. 17) be the foot of the object; A^ B^ C the three
points at which the angular elevation of the summit of the
object is the same ; then they must all be at the same distance
from 0. Let x be this common distance.
Let the angle AOB = d^ the angle AOC = <f). Measm-e
-BO, CA^ AB, and let their distances = r/, b, o, respectively ;
22 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
then a = 2x sin ^[d + ^),
b = 2x sin ^<^,
c = 2x sin ^0.
Eliminate ^, <^, from these three equations, then x will be
known and the distanee of the person from the object deter-
mined.
1850.
A person wishing to ascertain the distances between three
inaccessible objects -4, ^, (7, (fig. 18), places himself in a line
with A and 7i; he then measures the distances along which he
must walk in a direction at right angles to AB^ until A^ (7, and
-B, 0, respectively, are in a line with him, and also observes in
those positions their angular bearings: shew how he can find
the distances between A^ B, and C.
Let BE and BF, the measured distances, = d and e ; BEA
and BFA the obsei-ved angles = a and /3. Let the sides of the
triangle ABC = a, h, c, and BB = x.
Therefore tan BEA = ^4^ ,
a '
and tan BEB = ^ ,
a
.-. tan a = tRn{BEA - BEB),
X + C X
d d
^^x[x+cy
d'
cd
d'^ + x{x-{- c)
(1).
Similarly, tan^S = ~ —, ^ f2).
e- + x{x + c) ^ ^
From equations (1) and (2), a; and c are known, and thence
BAC = tan-^ -^ , and CBA = 180° - tan"^ - ,
and thence the distances a and h.
1851.] TRIGONOMETRY. 2^
1851.
. T<> /^ w sin a cos a
1. It tanp = ;— rr- ,
1 — /i sin a
shew that tan(a-/3) = (1 — w) tana.
■iiT 1 / ^N tana — tan/3
We have tan (a - /3) = :; : : — ^ ,
^ ^' I + tana tan/3 '
71 sin a cos a
tan a — ; ;— 5—
1 — n sm a
?? sin'' a
1 +
1 — n sni a
sin a
= tan a — n n sm a cos a,
cos a
sin a — «(sin a + cos a) sin a
cos a
= (1 — w) tan a.
2. Two triangles stand on the same base, determine in terms
of the base and of the tangents of the angles at the base, the
distance between the vertices of the triangles.
Let ABG^ ABC (fig. 19) be the two triangles. Let BC the
base = rt, and let the angles ABC^ A CB = B^ C, respectively,
and the angles A'BC, A' CB = j5', C Join AA', and let
AA' = ?•, then it is required to find the magnitude of r.
Draw ABj AD' pei-pendicular to the base. Then
r'= [AD-AD'f + DD'%
= {AD - AD'f + {AD cot B - AD cot By.
Now a = AD (cot B + cot (7),
also = AD (cot B' + cotC) ;
1 1
r' = a'
cotB + cotC cot 5' + cotO';
cot-B' cot 5
, / coti^' coti^ Y
"^ " Vcot B' + cot C cot B + cot 67
24 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
( / tan ^ tan C tan B' tan C"
(Vtan5 + tanC tani? + tanO
tanC tanC N^)*
,tani?' + tanC" tan 5 + tan (7/ j '
an expression of the required form.
25
CONIC SECTIONS.
1848.
1. Given the lengths of the axes of an ellipse, and the
positions of one focus, and of one point in the curve : give a
geometrical construction for finding the centre.
Let MN (fig. 20) be a line equal in length to the axis-minor.
With N as centre and a radius equal to the axis-major, describe
an arc of a circle. From 31 draw MO perpendicular to MN^
and cutting the arc in 0, MO will be equal to the distance
between the foci of the ellipse.
Produce 8P to Q (fig. 21) making ;S^^ equal to NO. With
P as centre, and PQ as radius, describe an arc of a circle, and
Avith S as centre, and radius equal to MO^ describe another arc ;
H the point of intersection of these arcs will be the other focus,
for /SlP, PH are together equal to the axis-major, and SII is
equal to the distance between the foci. If therefore we bisect
SH in (7, C will be the centre.
Since the arcs described from 6^, P as centres will in general
intersect in two points, it appears that there are two positions
which the centre may have.
2. P is any point in an ellipse (fig. 22), A A' its axis-major,
NP an ordinate to the point P; to any point Q in the curve
draw AQ^ A' Q^ meeting NP\n R and S-^ shew that
NR.NS = NP\
Draw the ordinate QM, then by similar triangles ANP^ AMQ,
XR : NA :: MQ : MA,
an<l by similar triangles A'NS, A'MQ,
NS: NA' :: MQ : MA\
26 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
therefore NE.NS : NA.NA :: MQ" : MA.MA ,
also NP^ : NA.NA' : : BC : A G\
therefore NR.N8 = NP\
3. FSp (fig. 23) is a focal chord of a parabola, BDr the
directrix meeting the axis in 2) ; Q is any point in the curve :
prove that if PQ, p Q produced meet the directrix in i?, r, half
the latus-rectum will be a mean proportional between DR., Dr.
Draw PjV, Qm pei'pendicular to the directrix, and join
/ST?, Sr^ SQ.
Then sinPi^^S' = sin P8R ^
PN
= dnPSR ^
= smPSR.&mPRN :
similarly sin QR8 = sin QSR.sin QRNj
.-. smP8R = s'ln Q8R,
.: Q8R=p8R:
similarly Q8r = P8r,
.'. R8r is a right angle,
.-. 8D' = DR.Dr,
or half the latus-rectiun is a mean proportional between DR., Dr.
1849.
1. Draw a parabola to touch a given circle in a given point,
so that its axis may touch the same circle in another given
point.
Let PQR (fig. 24) be the given circle, P the point in which
the parabola is to touch it, Q the point in which the axis is to
touch it. Draw PT a tangent to the circle at P, this will also
be a tangent to the parabola at P. Draw QT touching the
circle at (), and at the point P in the straight line PP, make the
1849.] CONIC SECTIONS. 27
angle TP8 equal to the angle PTQ] then S^ the intersection
of QT and FS, will be the focus of the parabola. Bisect TF in
K, and draw KA perpendicular to ST. A will be the vertex
of the parabola, and the vertex and focus being found, the cui^ve
may be constructed.
2. If a circle be described touching the axis-major of an
ellipse in one of the foci, and passing through one extremity
of the axis-minor, the scmiaxis-major will be a mean propor-
tional between the diameter of this circle and the semiaxis-
minor.
Let AA' (fig. 25) be the axis-major of the ellipse, S the
focus, C the centre, B the extremity of the axis-minor. De-
scribe the circle S£P touching AA' in >S', and passing through
By and draw the diameter SB.
Join SBy BB; then the angle SBB^ being in a semicircle, is
a right angle, also the angle SCB is a right angle. And the
angle BSP is equal to the angle SBCj therefore the triangles
BSBy SBC are similar. Hence
BC: SB:: SB: SB,
or SB is a mean proportional between SB and BC. But SB
is equal to the semiaxis-major ; therefore the semiaxis-major
is a mean proportional between the diameter of the circle and
the semiaxis-minor.
3. If ABj CBj two lines in an ellipse, not parallel to one
another, make equal angles with either axis ; the lines A C, BD
and ADy BC will also make equal angles with either axis.
Let A'B'C'D' (fig. 26) be the points of intersection of the
peq^endiculars to the axis-major through the points ABCD with
the auxiliaiy circle A'B'C'D'. Then it is evident that a line
joining any two of the above points as A'B' will intersect the
axis-major in the same point as AB does, and any two lines
joining the above points as A'B'.^ CD' will be equally inclined
to the axis-major, and therefore to either axis, if AB., CD arc so.
28 SOLUTloX.S OF SENATE-HOUSE PROBLEMS. [1850.
and vice versa: hence we have to prove that If A'B\ CD' are
equally inclmed to the axis-major LFE^ the lines A'C'^B'D'
and A'D\ B' C are so.
Now lAGL = L A EL + L B'A C\
and z D'HL = l UFL + l B'D' C ;
also /:AEL = I UFL,
and lB'AC = aB'D'C;
therefore lAGL = L D'HL = L B'HG,
or AC\ B'D' are equally inclined to LFE.
Again, z ALH = L LD'H + z LED',
and z B'KG = lKC G ^ lKGC ',
also lLD'H=lKC'G,
and lLED = lKGC]
therefore z ALE = L B'KG,
or AD' and 5 '6" are also equally inclined to LFE\ therefore
also AC, BD and AD, BC are equally inclined to either axis.
Q. E. D.
1850.
1. If from any point P of a circle, PC be drawn to the
centre (7, and a chord PQ be drawn parallel to the diameter
A CB, and bisected in R, shew that the locus of the intersection
of CP and AR is a parabola.
Let 0 (fig. 27) be the intersection of CP and AR. Draw
AM, CN pei-pendicular to AB, ONM parallel to AB. CN will
pass through R. Then
CO-.CP'.'. AO:AR,
:: OM:MN.
But CP=AC=MN,
therefore CO = OM,
and the locus of 0 is a parabola, of which C is the focus, AM
the directrix.
1850.] CONIC SECTIONS. 29
2. From the point P in the ellipse APB (fig. 28), lines are
drawn to A^ B, the extremities of the axis-major, and from
A, B^ Hues arc drawn perpendicular to AP^ BP; shew that the
locus of their intersection will be another ellipse, and find its
axes.
Let Q be the intersection of the lines pei'pendicular to
AP^ BP. Draw P2f, QN perpendicular to the axis-major, then
the triangles PBM, BQN ^yl\\ be similar, therefore
PM:BM::BN: QN',
similarly P3I : AM :: AN : QN,
therefore PM' : AM.BM : : AN.BN : ^.V''.
But if Z* (7 be the semiaxis-minor of the original ellipse,
PM' : AM.BM:: bC : AC\
therefore AN.BN : QN' ::hC'' : AC]
therefore the locus of Q is an ellipse whose axes are to one
another as IC : AC.
And if we draw AB\ BB' perpendieidar to Ah^ Bb^ we have
A (1^
^■^=#'
bC
which is one axis, the other is equal to -^-^ B'C or A C.
3. If two elUpses having the same major axes, can be
inscribed in a parallelogram, the foci of the ellipses will lie
in the comers of an equiangular parallelogram.
For it is evident that the centres of the ellipses must lie at the
point of intersection of the diagonals of the parallelogram, that
is, must be coincident, and their major axes are equal; there-
fore they will have a common auxiliary circle.
The lines joining the points of intersection of this circle and
the parallelogram, will, if the right points are joined, be pei'pen-
dicular to the sides of the parallelogram, and each of them will
contain two foci: hence the four foci will be at their points of
intersection, that is, at the comers of a parallelogram, equiangular
with the circumscribing parallelogram.
30 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [IBol-
4. If from the extremities of any diameter AB (fig. 29) of
an equilateral liyperbola, lines be drawn to any point P in the
curve, they will be equally inclined to the asymptotes.
From A and P draw the perpendiculars A C, AF, PE, PF,
on the asymptotes, AF and PF intersecting in Fj from B draw
BD perpendicular to the asjTuptote Oq. Then, since PQ = Bq.,
and the triangles QFP^ BDq are similar, they are also equal ;
therefore PE = Dq and QE = BD, therefore
AF=^ CO + EP=OD + Dq = Oq,
and PF = OE + AC = OE + BB = OE -\- EQ = OQ',
therefore the right-angled triangles APF, qQO are equal in
all respects, and the chords PA, PB equally inclined to the
asymptotes.
1851.
1. Given a pair of conjugate diameters of a conic section,
find geometrically the position of the principle diameters, (1)
in the case of the hyperbola, (2) in that of the ellipse.
Let PP', DU (fig. 30) be the given conjugate diameters
of an hj-perbola intersecting in C.
Join PD, PB'-, bisect them in E and F, draw GE, OF',
bisect the angle ECF by the line A' CA, and through C draw
BCB' pei'peudicular to ACA' ', these will be the principal di-
ameters required. For, by the property of the hyperbola,
CE, CF are the asjnnptotes, and AC A', BCB', to which they
are equaUy inclined, are the principal diameters.
2.* The solution of this part of the problem depends upon
the property of the ellipse, that if P (fig. 31) be any point in the
ellipse, and CR, CN, two lines at right angles to each other, cut
the straight line PRN in the pomts R, N, such that PN is equal
to the semiaxis-major, and PR to the semiaxis-minor, CR and
CN will be the directions of the principal axes.
* For this solution the authors are indebted to the kindness of the Mode-
rator, Mr. Gaskin.
1851.] CONIC SECTIONS. 31
Let CP, CD be the given semi-conjugate diameters; draw
Pi^ pei-pcndlcular to CD'^ make FK equal to CP; upon CK as
diameter describe the circle CFK\ tlii'ough its centre 0 draw
PRN'^ join CP, CN'. these will be the directions of the principal
axes.
For PF. CD = PF.PK = PR.PN = A C.BC,
and CP' + CD' = CP'' + PK' = 2 CO' + 2P0' = 2 OE' + 2 0P\
= PR + P^ {Euc. II. 10) = J.0"^ + PC'^
therefore PN = AC, PR = BC,
and OP, CW are the du*ections of the principal axes.
{ 32 )
STATICS.
1849.
1. Two forces F and F'^ acting In the diagonals of a paral-
lelogram, keep it at rest in such a position that one of its edges
is horizontal; shew that i^seca = i^'seca' = TFcosec(a + a'),
where W is the weight of the parallelogram, a and a! the angles
between its diagonals and the horizontal side.
Let AB (fig. 32) be the horizontal side of the parallelogram.
In order to preserve equilibrium, the directions of the forces
F^ F' must meet in G^ the centre of gravity. Hence, by the
triangle of force,
F _ F' _ W
smBG W ~ &\nA G W ~ sin^ GB '
F F' W
or
cosa cosa sm(a + a)
therefore i^seca = i^'seca' = TFcosec(a -I- a').
2. A cubical box is half-filled with water, and placed upon
a rough rectangular board ; if the board be slowly inclined to the
horizon, determine whether the box will slide dowTi or topple
over.
Let fi = the coefficient of friction.
Then the box would begin to slide when the inclination of
the board to the horizon = tan~^yLt.
It would begin to topple when the inclination = jtt.
Therefore it will begin to slide or topple over, according as
/Lt < or > 1.
1850.
1, A heavy body is supported in a given position by means
of a string which is fastened to two given points in the body.
1850.] sTATifs. •};^
and then passes over a sjnooth peg: find the k-ngtli ot" the
strmg.
Let G (fig. 33) be the given position of the centre of gravity,
A and B those of the points of support. The position of tlie
peg P is determined by the conditions that it must lie in the
vertical tlu'ough G^ and that the angles APG, BPG must bi;
equal, each = 6 suppose.
luQt AG = a, BG = b, lA GP = a, L BGP = /3 ; then
PG miPAG sinf^ + a)
AG sin APG sin6>
= cos a + sin a cot^
PG
similarly -^^ — cosyS + sin^cot^,
, ^ cosa + sina cot^ BG b
thereiore y^ -. — ^ 7: = -j-f, = - ,
cosp + smp cota AG a
whence 6 is known, and length of the string
= AP^BP,
sin a sin /3 , . ,
= -. — 7i a + -. — - b is known,
sma sma
2. Two spheres are supported by strings attached to a given
point, and rest against one another: find the tensions of the
strings.
Let -4, B^ (fig. 34) be the centres of the spheres, and C the
peg. Then, if the spheres are smooth, the strings must lie in
the lines CA^ CB; hence the parts of the triangle ABC ai-e
known. To determine its position.
Let G be the centre of gravity of the spheres, CG must be
vertical. Let W^, W^ be the weights of the spheres A and B,
therefore AG : BG :: W^ : W^-,
and if LACG = e,
smjC- 6) _ BG sin^ _ TF, sini? ^
siiT^ "" ATCf&mA ~ it; sin ^ '
1 /• • /> /I n ^^^, sin 5
therefore smC cot a — cosO = ttt • — 7 1
T^.^sm^ '
whence 6 is known.
34 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
Let T,, T^ be the tensions of the strings which support
A and B respectively; therefore resolving the forces on the
sphere A perpendicular to AB^
T, sin^ - PF; sin(^ + ^) = 0,
_ sin(^ + ^) ,.^ .
or
sin -4
and similarly 1\ = t^^ ^ TT,,
_sin(^ + 6>)
~ sin 5 «'
whence T^ and T^ are known.
3. A cone of given weight W (fig. 35) is placed with its
base on a smooth inclined plane, and supported by a weight
W\ w^hich hangs by a string fastened to the vertex of the
cone, and passing over a pully in the inclined plane at the
same height as the vertex. Find the angle of the cone when
the ratio of the weights is such that a small mcrease of W would
cause the cone to turn about the highest point of the base, as
well as slide.
Let a = the angle of the plane,
6 = the half-angle of the cone.
Since the resolved parts, along the plane, of the tension W of
the string and the weight W just balance, we have
PTsina = T'T'cosa (1) ;
and' because the moments of the same forces about B are also
equal,
Wmia.lAC+ Wcosa.BC = W cosa.AC - W sma.BC,
W cosa.f^O = ( W ^ + W sin a") BC, from (1),
or BC = f sina cosa A C,
or tan^ = # sin 2a.
1851.] STATICS. 35
1851.
1. A cone whose semi-vertical angle is tan"* -j is enclosed
in the circumscribing spherical surface, shew that it will rest in
any position.
Let ABC (fig. 35) represent a section of the cone made by
a plane through its axis. Divide the axis AD in (r, so that
GD = iADj then G will be the centre of gravity of the cone.
Join BGj then
BG' = BD"' + DG\
But
i'AnBAD = ^^,
therefore
BD' = \AD\
and
DG' = ^AD\
therefore
BG' = ^AD\
and
BG = IAD,
^AG;
therefore G is the centre of the circiunscribing sphere.
Hence it appears that the height of the centre of gravity of
the cone will be the same in whatever position it be placed,
therefore it will rest in any position.
2. A string ABCDEP (fig. 37) is attached to the centre
A^ of a pully Avhose radius is ?•, it then passes over a fixed point
B^ and under the pully, which it touches in the points C and D ;
it afterwards passes over a fixed point E^ and has a weight P
attached to its extremity ; BE is horizontal and = — , and DE
is vertical : shew that if the system be in equilibrimn the weiglit
bP
of the pully is — , and find the distance AB.
Let W be the weight of the pully, and let ^, ^ denote the
respective inclinations of AB^ BC to the horizon. The tension
of the string will be thi'oughout = P; hence resolving horizontally
d2
36 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
and vertically,
Pcos^ - Pcos<^ = 0 (1),
P(l + sln^ + sin</)) = TF. (2).
Again, AB = r cosec(^ + ^),
5?'
therefore EB = — = r {I + cosec[6 + </>) cos^},
therefore |sm(^ + 4>) = coaO (3).
By (1) e = </,,
and by (3) fsm2^ = cos^,
therefore sin^ = ],
5P
therefore by (2) W = — .
Also AB = rcosec(^ + ^),
r
^ sm2^ '
r
- -^
"3:7*'
which gives the distance AB.
( 37 )
DYNAMICS.
1848.
1. Two bodies acted on by gravity are projected obliquely
from two given points in given directions and with given
velocities: determine their position when their distance is the
least possible.
Let the bodies A^ B be projected from the points A^ B
(fig. 38) in directions AC^ BC intersecting in (7, and with
velocities proportional to CE and CD] upon both the bodies
impress a velocity CE equal and opposite to ^'s velocity, and
suppose gravity not to act, the relative motion of A and B will
not be affected by either of these circumstances ; but A will now
be reduced to rest, and B will move in a direction BG parallel
to the diagonal CF of the parallelogram on CE^ CD. From A
draw AG perpendicular to BG^ AG will be the shortest possible
distance between A and B] and A and B will be at that dis-
tance at the time [t] after the instant of projection that it takes
a body animated with the velocity CF to describe the space BG^
a kuo\\Ti time therefore. Let AH and BK be the spaces due to
^'s and -B's velocity of projection in time t. Through H and K
draw HL and KM^ each equal to the space due to gravity in the
tunc t ; L and M are the positions required.
2. A railway train is going smoothly along a curve of
500 yards' radius at the rate of 30 miles an hour ; find at what
angle a plumb-line hanging in one of the carriages will be in-
clined to the vertical.
Let a denote the inclination of the plumb-line to the vertical,
38 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
CO the angular velocity of tlie train per second, r the radius of
the cur\'e.
Then the weight at the end of the plumb-line may be con-
sidered to be in oquilibrium under the action of the centrifugal
force, gravity, and the tension of the string.
Hence (fig. 39), by the triangle of forces.
sin (tt — a) sin (^tt + a) '
coS' g
or
sm a cos a
tana = .
9
AT „. .r^. 30x5280 44
Now g = 32.2, . = 1500, a> = ^^^^-^^^^^0 = 1500 '
(44
.'. tana =
1500 X 32.2 '
^ (44F
48300 '
- 4^4
~ 12075 '
which gives the inclination to the vertical.
3. A nmnber of balls of given elasticity A, B, C are
placed in a line ; A is projected with a given velocity so as
to impinge on B] B then impinges on (7, and so on: find the
masses of the balls B^ C , in order that each of the balls
A, B, C may be at rest after impinging on the next;
and find the velocity of the «*'' ball after its impact with the
(n - 1)'".
Let m : 1 be the ratio of the mass of w* ball to that of the
{n— 1)'^', then the ratio of the velocity of the {n— ly^ ball after
impact to its velocity before, would be, if the balls were inelastic,
_ 1
~ 1 + m '
1848.] DYNAMICS. 39
Since they are elastic, the ratio is
= 1 - (1+ .) f 1 - ^
= 1 - (1 + e)
1 + mj '
m
1 + m
and since the (w — 1)"* ball is thus brought to rest, this must = 0,
.-. 1 + - = 1 + e,
m
and ni = - ,
e
so that the masses of the balls from a geometrical progression,
whose common ratio is -, and the ratio of the velocity of the
I -\- e
n*^ ball after impact to that of the (n— lY^ before = ;; = e :
therefore if V be the initial velocity of A, velocity of n^^ ball
after impact = e"~' V.
4. An imperfectly elastic ball is projected in a given direction
within a fixed horizontal hoop, so as to go on rebounding from
the surface of the hoop ; find the limit to which the velocity of
the ball will approach, and shew that it will attain this limit at
the end of a finite time.
Let e be the modulus of elasticity, F, V^ F^ the velocities
of the ball before the first, second, (n— 1)'^ impacts, 6, ^,...^„
the successive angles of incidence. Then
Fj cos 0^ = e Fcos ^,
F,8in^, = Fsin^,
.-. F/ = F^(sin'^^ + e"'cos^^),
= sin''^(l +e:'coi'e)V':
shuilarly F/ = sin'^6», {I + e' cof''^,) V;\
1 1
But sin'^. =
1 +cof^, 1 + rVot'^'
40 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1B49.
therefore T^ = sin''6> (1 + c? cof^^J V%
and similarly it may be shewn that
T7 = sin-^^(1 + e'^"cof^^)F'';
hence when n is indefinitely increased,
V = Fsin(9,
the limit to which the velocity of the ball approaches.
Now the distances between the successive points of incidence
are 2rcos^ , 2?'Cos^.^ r being the radius of the circle; there-
fore the times of describing these spaces are
2?-cos^, 2rcos^„ . ,
— 1^- ' , — T^— ^ respectively,
1 2
cot 6', 1 , cot^„ 1
= 2r
:i + cof^j* TV (i + cof^^j* f;
('cot^ (1 + e^cot-^ji 1
H-e'^cot'6')4 sin (9 F'
e'cot^ (l+e*cot'^)* 1
2r
:i +e*cot'6')4 sin^ F" '
2r cos^ 2r jjcos^
"" F^ii^' F^ dK^ '
therefore the ball will attain its terminal velocity, after the time
2r cos^ / 2 s \
2r cos^ e
V sin'^ I -e'
1849.
1. A body is projected from a given point in a horizontal
direction with a given velocity, and moves upon an inclined
plane passing through the point. If the inclination of the plane
vary, find the locus of the directrix of the parabola which the
body describes.
Let a be the inclination of the plane to the horizon ; F the
velocity of projection ; / the latus-rectum of the parabola de-
1849.] DYNAMICS. 41
scribed ; therefore
j^r sina '
and I sill a = .
9
But J? sin a is the height of the directrix above the given
point of projection ; therefore this height is constant, and the
locus of the directrix is a horizontal plane at a distance — above
2g
the given point.
2. An imperfectly elastic ball A lies on a billiard-table,
deteniiine the direction In which an equal ball B must strike
it in order that they may impinge upon a side of the table
at equal given angles.
The impact must be oblique and the impulse take place in
the direction in which A is to go off. This direction makes
with the side of the table the given angle a : let ^ be the angle
which ^'s direction before impact makes this direction.
V = B^s velocity before impact,
e = the modulus of elasticity.
5's velocity V sin 0 pei'pendicular to the direction of the Impulse
will be unaltered by it: if there were no elasticity, its velocity
in direction of the impulse after impact would be ^Fcos^, since
the balls are equal, and the impulse ^MJ^cosd: hence the actual
impulse will be ^(1+e) MVcosd, and the actual velocity in its
direction after impact P^cos^ — ^{1 + e) Fcos ^ or ^(1 — e) Fcos^.
Let <f) = the angle which ^'s direction after impact makes
with the direction of the impulse,
tan(/) = yy- r ^ = j— r taU ^.
■^(l-e)cos^ i{l-(')
But (f> = 2a,
.-. tan^ = ^(1 -e) tan 2a,
whence 6 is known.
42 .SOLUTIUXS OF SENATE-HOUSE PliOBLEMS, [1849.
3. A bead running upon a fine thread, the extremities of
which are fixed, describes an ellipse in a plane passing through
the extremities, under the action of no external force; prove
that the tension of the thread for any given position of the bead
is inversely proportional to the square of the conjugate diameter.
Let the bead be at the point P of the ellipse.
Since the tension of the string is the same tkroughout, the
resultant force on the bead will bisect the angle SPH^ and
therefore be nonnal to the elliptic path. Consequently, as no
force acts upon the bead in the direction of its motion, its
velocity will be uniform. Now, considering the bead as moving,
for the instant, in the circle of curvature at the point P, nomial
J, vel.' 1 .,,...
torce Gc — 3 — 7: cc — j — ^ , smce the velocity is mii-
rad. ot cui'v. rad. 01 curv. •'
form : but radius of curvature gc CD\ therefore normal force
1
Now, adopting the usual notation,
tension of the string : nomial force -.-.PE: PFr. CD.AGxCD.PF,
'.iGB-.BC,
therefore tension of the string cc -^^ .
4. The centres of two equal spheres (elasticity e, radius r,)
move in opposite directions in a circle (radius B) about a centre
of force vaiying inversely as the square of the distance ; deter-
mine the motion of the spheres after they have impinged, sup-
posing that e = ^ 5 ; and prove that the latus-rectum of the
conic section described after the second impact will be 2e^i?.
Let 0 (fig. 40) be the centre of force ; (7, C the centres of
the spheres. Draw OPQ perpendicular to CC\ such that CQ
is perpendicular to 00, and consequently C Q to OC. Then
if CQ represent in magnitude and direction the velocity of the
sphere 0 before impact, CPj PQ will represent its resolved parts
in directions CP^ PQ. Now draw QB perpendicular to 0'^,
1850.] DYNAMICS. 43
meeting CP in B: the triangle QPR is evidently similar to
G'PQ^ and therefore to CPQ. Hence
BP:PQ::PQ: CP,
.: PPiCP:: PQ' : CP' :: CP' : OP' :: r' : B' - r' :: e : I.
Consequently BP represents in magnitude and direction the
resolved part, pei'pendicular to OQ, of (7's velocity after impact.
The velocity PQ remains unaltered by the impact ; therefore
the diagonal of the parallelogram PBQ drawn through P will
represent in magnitude, and be parallel to the direction, of the
whole velocity of C after impact. Now this diagonal makes
with PQ an angle equal to BQP or COP or COP, and is
therefore parallel to OC. Hence after impact the centres of
the spheres will move directly from the centre 0 in the lines
OC, OC. They will evidently return to the same positions
C and C, and there impinge a second time.
For the velocity of C after the second impact it is sufficient
to obser\'e that the velocity along OP will be unchanged, while
that perpendicular to OP will be again diminished in the ratio
of ?: 1. Let PB' = e.PB. Through C draw CS equal and
parallel to QB' ; join OS. Therefore the latera-recta of the
first and third orbits will be to one another as (triangle OCQY
: (triangle OSCf, since these triangles represent upon equal
scales, half the product velocity x pei*pendicular on the tangent ;
and we may shew that (triangle OCQf : (triangle OSC)' :: 1 : e* ;
and the latus-rectum of the first or circular orbit is 2B. There-
fore that of the third is 2e^B.
1850.
1. Shew that it is possible to project a ball on a smooth
billiard-table from a given point in an infinite number of
directions, so as, after striking all the sides in order once or
oftener, to hit another given point ; but that this number is
limited if it have to return to the point from which it was
projected.
Let P (fig. 41) be the point of the table from which the ball
is projected, PQBSTU its course once round the table. 7?.S'
44 S(.>LIT1U.NS OF SENATE-HUUSE I'liUBLEMS. [1850.
may be sliewn to be parallel to QP-^ and if the elasticity be
perfect, equidistant with it from the line AD drawn through the
comer A of the table parallel to either of them. For the angle
SUB = angle QRA = 90° - BQA^^OT - PQD, therefore RS
is parallel to QP. Also if QR intersect AD in F,
RF'.QF:: RA sin RAF: QA smQAF,
:: RA s'mBRS : QA sinDQP,
:: RA sm ARF: QA sinAQF.
But RA sm ARF = QA sin A QF,
therefore RF = QF^
and RSj QP are equidistant from AD.
Similarly, RS and TU are equidistant from CE. Through
P draw VDEPU pei'pendicular to the parallel lines. Then
VD = DP and VE = EU, therefore
PU= DE+ EU- DP=DE+ EV - DV= 2DE.
The same equation, PU = 2DE, holds whether P and U be on
the same side of D and E or on opposite sides of either or both.
Hence it is evident that by choosing the direction PQ rightly
we may make the ball hit the second given point, through
■which the line TU will pass, after striliing all the sides once :
and by lessening DE or projecting the ball more nearly in the
direction of the diagonal CA, we may make it strike the second
point after striking all the sides twice, when PU will = ADEy
and so on ; there being thus an infinite number of directions of
projection each more nearly parallel to the diagonal CA than
the preceding, which will cause the ball to hit the second given
point after striking all the sides once, twice, &c., respectively.
If, however, the ball have to return to the point of projection,
we must have DE = 0, or the direction of projection parallel to
either diagonal; there being thus two directions and their op-
positcs, or four directions in all, which will bring the ball back
to its point of projection. Through this point it will pass after
making each round of the table.
DYNAMICS. 45
1851.
1. A body of given elasticity e is projected along a hori-
zontal plane from the middle point of one of the sides of an
isosceles right-angled triangle, so as, after reflexion at the
hypothenuse and remaining side, to return to the same point ;
shew that the cotangents of the angles of reflexion are e + 1
and e + 2, respectively.
Let ABC (fig. 42) be the triangle, right-angled at -4 ; i> the
middle point of AB the point of projection : AD — DB = a.
Let e be the modulus of elasticity. Draw DEF perpendicular
to BC^ making EF = e.ED: draw FGH perpendicular to ylC,
making GH = e.GF: di'aw HD, LF, KB: DEL will be the
path of the body.
The angle of reflexion at -ff" = 90° — LEG = 6 suppose,
L= LEG =</)
Now BD = «, .-. BE= ^, =-. EL
' 2* e '
.-. /6'=2a.2i- [l+e) BE,
= 2a.2i-(l + .)-j,
.-. AG = 2a- CG= (l+e)K
and FI = ae, IG= CG = {S- e) ^a,
.-. HG = e.FG = e (3 + e) ^a,
,._GH_ GH+AD_ l + ^e(3 + g)a
.-. COt(^_ ^^- ^^ - ^(i_^g)„
^(2 + e)(l-he)
l+e '
= 2+e (1).
46 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Again, cot 0 = tan LKC = tau (45° + KFI) ,
1 + tarn KFI
1 - ts^u KFI'
1 + e t&n LEG
~ 1-e tsmLEG '
e
'-.le
2 + 2e
~ 2 '
= 1 + e
1)
and
(2)
give
the
results required.
(2),
2. If a heavy body be projected in a direction inclined to
the horizon, shew that the time of moving between two points
at the extremities of a focal chord of the parabolic path is pro-
portional to the product of the velocities of the body at the two
points.
Let S (fig. 43) be the focus of the parabola, PSj) the focal
chord, mKM the directrix, KAS the axis ; draw PiV, pn per-
pendicular to the axis, PM^ jim to the directrix. Then, since
the body is acted on by no horizontal force, its horizontal
velocity will be constant, and therefore the time of moving
from P to 10 will be proportional to PN+pn. Also (velocity)''' at
P =2g.PM^ (velocity)'"' at^ = 1g.im\ consequently the problem
is solved if we can shew that {PN -Vpi)^ \ PM.pm is a constant
ratio.
Now PM = 8K+ SP cobPSN = 2AS + PM cosPSN,
.-. PM (1 - co&PSN) = 2AS.
Similarly, pm (1 + cos jpSn) = pn (1 + cosPSN) = 2A8j
.-. PM.pm Bm'PSN = ^AS' (1).
1851.] DYNAMICS. 47
smPSN
Also PN = SF smFSN = PM smPSN = 2AS
I- cos PSN
from above.
Similarly, pn = 2Ab nci\T== 2 AS- ttcptt'-)
•^ ' -^ 1 + cosPSN 1 + cosP/SiV
TiAT . A o sinP/SiV , . ^ 1
.-. PX+mi = AAS , 2novr= ^^'S^ . noAr;
^ 1 - cos" PSN sm PSN ^
.-. (P.V+^n)^ sm-'P/S'^Vzz^ 16^^^ (2).
From (1) and (2),
{PN-\-p7iY : PM.pn :: 4 : 1, a constant ratio. Q. e. d.
48
NEWTON.
1849.
1. The circle described through two points of an equiangular
spiral and the point of intersection of the tangents at those
points will pass through the pole. Prove this, and apply tlie
proposition to shew that the curvature at any point of an equi-
angular spiral varies inversely as the distance of the point from
the pole.
In the equiangular spiral the tangent is inclined at a constant
angle to the radius vector ; hence in (fig. 44) if P^ T, P^ T be the
tangents at the points P^P^^ S the pole of the equiangular
spiral P^P^,
SP^T+ 8PJ=TT,
and a circle can be described about the quadrilateral SP^TP^^
or a circle passing through P„ T, P^, will also pass the pole S.
Suppose P^P^ to be indefinitely near to each other, then P^ T
ultimately becomes equal to i^^ T, since the triangles SP^T^ STP^
ultimately become similar and equal. Produce SP^ to meet P^, T
in B, and draw P^R' perpendicular P^T; then, ultimately,
P^E' = P^EsmSP^T.
Now diameter of curvature at P^
= hmit ^ = ^^p-^ hmit -^^ ,
= g.^^py limit ^'^ by the above property,
2P,S
sin SP^T'
or the curvatm'e at P, varies inversely as /SP,, since SP^ T is
a constant angle.
NCWToX. 49
1850.
1. If any number of particles be movhig in an tllipse about
a force in the centre, and the force suddenly cease to act, shew
/ 1 \'^ . .
that after the lapse of I r— ] part of the period of a complete
revolution, all the particles will be in a similar, concentric, and
similarly situated ellipse.
The velocity at any point P (fig. 45) of the orbit = fM^CD^
2ir /I \-^
and the time of revolution — ; therefore after the [ — ) part of
a revolution, each particle wiU have described a space PP' equal
and parallel to CD. If therefore we complete the parallelogram
PCD, P' will be its angular point.
Join CP' meeting the ellipse in Q, and PD in V. Then, by
a known property of the ellipse,
CV.CF = CQ\
and CP' = 2Cr;
.-. CF'' = 2GQ%
and CP' = 2iCQ;
therefore all the particles are in a concentric, similar, and
similarly situated ellipse.
2. Two perfectly clastic balls are moving in concentric
circvdar tubes in opposite directions and with velocities pro-
portional to the radii : at an instant when they are in the same
diameter and on opposite sides of the centre, the tubes are
removed and the balls move in ellipses mider the action of
a force of attraction in the common centre of the circles vaiying
inversely as the square of the distance. After one has per-
formed in its orbit a complete revolution and the other a
revolution and a half, a direct collision takes place between
the balls and they interchange orbits. Find the relation between
the radii of the circles and between the masses of the balls.
Let r^, r,^ be the radii of the circles. Then the greatest and
least distances in the two orbits will be ?•,, r^ in the first and
r„, r^ in the second, where i\ has to be determined.
E
50 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
Now let hji.^ be the values of h in the two orbits, therefore
Aj vel. in circle rad. i\ x r^
\ ~ r^^r^ '
= ^ (').
a
since the velocities are proportional to the radii.
h '■' latus-rcctum in first orbit
^^'^ t'= second...'
2
= ^'i + ^'^ = !^ . ''^ + '"3
' •/ a _ 3 1^2
^2 + ^
. !\! = !\ ^'2 + ^3 ^2)
•• ^4 ,. • ;. _|_ ,. ^ ''•
'2 '31' 2
Also the periodic time in the first orbit = f that in the second, or
-m^(!f '«)^
therefore, from (2) and (3),
r; _ /3M r.
r* \2J r.
r, _ /3M r.
and, from (3),
-©*-(i)H-(i)n-(i)-"'
1851.] XEVVTOX. ')l
r.
the equation tor finding — . This equation has only one positive
root, and that less than 1, as it ought to be, n suppose.
To find the relation between the masses m^ and nr, of the
balls in the greater and less orbit respeetively. Let i\ and ?'.^
be tlieii' velocities before impact ; their velocities after impact
will be I'., and i\ respectively, m^ and m,, both moving after
impact in the same direction as m^ the greater did before
impact. Hence, since the elasticity is perfect, momentum lost
by 7»j = the whole momentum lost and gained by vn^, or
and since the balls are at the same distance from the centre of
force, and moving in opposite directions,
^ = ^ = i' by (1)
2 2 2
1
V
-^ + 1
-1-1
^2
1 + )l'
the required relation between the masses.
1851.
1. If a body describe an ellipse round a centre of force in
the focus, shew that the sum of the reciprocals of the squares of
the velocities at the extremities of any chord passing through
the other focus is constant.
Let PHp (fig. 46) be the chord thi'ough II. Draw the pi-r-
pendiculars SY^ St/, HZ, Hz, to the tangents at those points:
join SP, Sj).
e2
52 SoLlTlONb OF SliNATE-lluUSl-: PROISLEMS. [1851.
Then, by a known property,
1 14
^^^ + T7- = T [L the latus-rectum),
JJ.F lip L
2AC , 2AC ,_B.^C
SP Sp SAC .
SY Sy SAC ^
or •.• SY.HZ^ SyJIz = BC'\
SY^ Sjf_^SAM
BC "^ BC ~ L '
or SY^ + Sy^ is constant,
and the velocities at -P,^?, are inversely proportional to 8Y^ Sy,
therefore sum of the squares of the reciprocals of the velocities
at Pp are constant.
Cor. It may also be shewn that the sum of the squares of
the velocities at the extremities of any chord passing thi'ough
the centre of force is constant.
For we have shewn that
SY Sy , ,
^^-^ + -~ = constant,
HZ Hz '
SY.HZ Sy.Hz ^ ^
••• ^^2 + jj^^ = constant,
or BC'^ [iTt'' "*" 77^ ) ~ constant,
'*' TiZ' "*" 'Hz^ ^ constant,
or, taking H as the centre of force, the siun of the squares of the
velocities at the extremities of any chord passing through the
centre of force is constant.*
* TTiis corollary was set as a problem in 1848.
I
( 53 )
HYDROSTATICS.
1848.
1. An inverted vessel formed of a substance which is heavier
than water contains enough of air to make it float : prove that
if it be pushed down througli a certain space, it will be in a
position of unstable equilibrium ; and deteraiine the space in
question.
When the vessel is floating partly immersed, the weight of
the water displaced is equal to the weight of the vessel and of
the air it contains. If the vessel be now pushed down, the
water displaced, and therefore the upward pressure on the vessel,
will be increased till the vessel is wholly immersed ; as the
vessel is now pushed down further the water displaced becomes
less on account of the compression of the air in the vessel, till
it comes into such a position that the weight of the water dis-
placed is only equal to the weight of the vessel and the air it
contains. This will be a position of equilibrium ; and the equi-
libi-ium will be mistable, for accordingly as it is a little above
or a little below tliis position, the weight of the water displaced
will be greater or less than that of the vessel and the air it
contains.
This explanation applies to a vessel of a cylindrical form ;
if, however, it is smaller at the top than the bottom it may come
into the position of unstable equilibrium before it is wholly
unmersed. To find how far the vessel must be displaced so as
to come into this position.
Let W be the weight of the vessel, V its volume ; a and j-
the altitude of the column of air in the vessel in the positions
54 SOIA'TIONS (IF SENATE-HOUSE PROBLEMS. [1848.
of Stable and luistablc equilibrium, V'V" its volumes in those
positions, 1/ the depth of its lower surface below the surface of
the water in the latter position: a the density of the water,
p, p those of tlie air in the two positions, jjy p its pressures in the
same positions.
Then, equating the weight of the fluid displaced and of the
vessel and air contamed,
^TV= W+<jV"p' (1).
Also equating the upward pressure of the water and downward
pressm-e of the air at their common surface,
!/(^!/ =P' (2)-
Also, since pressm'e and density vary inversely as volume,
^ = ^ = {; (3).
Again, when the form of the vessel is known, V and V" will be
known in tenns of a and .r. Hence equations (1), (2), and (3),
will be sufficient for the determination of x and ?/, as well as the
other unknown quantities, viz. ^j', p', and V". Hence y — x^ or
the depth of upper surface of the air below that of the Avater, is
knoAvn, and added to the difference of the heights of the same
two surfaces in the original position of equilibrium, gives the
space through which the vessel must be depressed.
2. A uniform piston, terminated by a plane of area A^ per-
pendicular to its side, is inserted into an orifice in a vessel
containing fluid ; prove that the work done in gently pushing
in the piston through a small space s is ultimately equal to the
work done in lifting a portion of the fluid of volume As through
a height equal to the depth of the centre of gravity of the plane
below the surface of the fluid.
If the space s be indefinitely small, the pressure on each
element of the piston will l)o unaltered by the change of the
pistoirs^ position.
1849.] HYDROSTATICS. 55
Hence, i( s be indefinitely small,
work done = product of whole pressure on area A x s^ the space
through which it is moved perpendicular to itself,
= pressure at depth z of the centre of gravity of
A X area A x Sy
= gpzA^y
= ffpAs.z,
= weight of volume As of the fluid x 2,
= work done in raising the volume As through the
space z.
3. Two equal slender rods ABj AC^ moveable about a hinge
at Ay and connected by a string BC^ rest with the angle A
immersed m a given fluid ; determine the tension of the string
BC.
Let T = tension of the string,
w = weight of each rod,
2a = its length,
2l = the length of the part immersed,
a = its inclination to the horizon.
Then the rod is kept at rest by its weight, the tension of the
string, the action at the hinge and the fluid pressures which
have for resultant a vertical upward pressure w acting at a
distance I from A.
Hence, taking moment about A,
10 .a cos a — w.l cos a — T.2a sin a = 0,
1 = -— — coicnAv
2a
is the required tension.
1849.
A body floats in a mixture of two given fluids with a volume
A immersed; one half of the mixture being removed, and its
place supplied by an equal quantity of the lighter fluid, the
same bodv floats Avith a volume A + B immersed. Determine
56 SOLUTIONS OK SENATE-HOUSE PROBLEMS. [1850.
the ratio ot" the quantities of fluid in the original mixture, sup-
posing the volume of the mixture to be equal to the sum of the
volumes of the eomponcnt fluids.
Explain the result when the densities of the fluids are as
^ + ^ to ^ - ^.
Let F, V be the original volumes of the fluids ; cr, a' their
specific gravities. Their volumes in the second mixture will be
^V and ^V + ^{V + V) or V + ^T'; the specific gravities of
the mixtures will be
hence, if W be the weight of the body,
PWFV
n -A ^.^^. ,
also = [A + H] p _^ y )
.-. A[Vct+V'<t') = {A+B) [iFo- + (F + iF)(7'},
.-. {AcT-^{A + B)a-^{A+B)a'} V^Ba'V,
V 2B(r'
or
F ' [A-B] (7-{A + B)a"
the required ratio.
If tlie densities, and tlierefore the specific gravities, are as
A + B io A — B, F' = 0, shewing that the fluid cannot be a
mixture of fluids of diiferent specific gravities ; in fact, the con-
ditions of the problem then become impossible.
1850.
1. A conical vessel containing a given quantity of fluid has Its
axis vertical, and another cone with the same vertical angle is
placed to float in the fluid with its vertex downwards ; find how
much the fluid will rise in consequence.
Let h be the depth of the original cone of fluid, k the depth
to which the vertex of the floating cone will sink ; A- is known
from the specific gravities of the fluid and floating cone, z the
1850.] HYDROSTATICS. 57
height thi'ough which the fluid will rise. Then
volume of the cone height /<! + s {h-\- zf
h " 1i' '
, volume of the cone height k k
and 7 = T? •
h h^
Therefore, subtracting,
volume of the fluid after the cone Is put Into It _ (A + zf — k^
its original volume K^ '
= 1,
since the quantity of fluid is unaltered ; therefore
{h + zf -k' = F
and z = {h' + k'Y - h
is the required space.
2. A hollow cylinder containing air Is fitted with an air-
tight piston which, when the cylinder is placed vertically, is at
a given height above the base ; the cylinder being now inverted
and placed vertically in a fluid, sinks partly below the smiace ;
find the position of equilibrium.
Let J) be the pressure of the air in the cylinder before the
cylinder Is inverted, and which Is knoAvn from the given height
(/«) of the piston above the base : p is the pressure due to the
weight of the piston and atmosphere, 11 the atmospheric pres-
sure, to the weight of the cylinder and piston, z the depth
below the sm*face of the fluid of the piston In the position of
equilibrium, i/ the distance of the piston from the base of the
cylinder, p, p the densities of the fluid and uncompressed air ;
then, for the equilibrium of the piston,
fluid pressure fi'ora beneath = pressure due to the weight of the
piston + the pressure of the air in the inverted cylinder,
ov c/pz + U =2^-U + -i> (1).
58 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Also ic + weight of the air iii the eyliiider
= weight of the fluid displaced,
.-. w+gp'V = fjpj^, V (2),
where Via the volume and // the height of the cylinder.
From equation (2) z is known, and thence y from (1).
1851.
1. A hollow cone floats m a fluid with its vertex upwards
and axis vertical ; determine the density of the air in the hollow
cone.
Let 2^ be the pressiu'e of the air in the cone, IT that of the
atmosphere, tv the weight of the cone, h its height, z the height
of the cone of compressed air, y the depth of its base below the
surface of the fluid, /?, p' the densities of the fluid and uncom-
pressed air.
Then, equating the pressures at the common smface of the
air and fluid,
gpy +'n = pressure of the compressed air,
=i> = -n (1).
Also 10 + weight of the air in the cylinder
= weight of the fluid displaced,
,TT z^ — (z — vY _- , ,
or to + gpV=gp Jf-^ ^ (2),
where V is the volmne of the cone.
From equations (1) and (2) z and y are known, and thence j-j,
and the required density.
( '^9 )
OPTICS.
1848.
1. If Q^ q (fig. 47) be two points in the radius of a spherical
reflecting surface whose centre is E^ such that EQ : Eq :: sine
of the angle of incidence : sine of the angle of refraction, de-
termine geometrically the position of the point P, so that a ray-
proceeding from Q and incident upon the surface at P may
after refraction proceed from q.
Bisect Qq in m, and thi'ough m draw mF perpendicular to
EQ meeting the circle in P; Pwill be the point required. For
if we join FE^ Fq^ FQ, we have
sin^P^ : sin FQE:: QE : FE,
and sm^P^ : sinP<^^ :: qE : FE.
Now sin FQE = sin FqE, •. • z FQE = LFqQ,
.-. smEFQ : sinJ;P^ :: EQ : Eq :: fM : I
by the question, therefore the ray QF after refraction at P will
proceed as if from q.
2. If a ray of light, after being reflected any number of
times in one plane, at any nimiber of plane sm'faces, retmii on
its fonner course, prove that the same will be true of any ray
parallel to the foi'raer which is reflected at the same surfaces
in the same order, provided the number of reflections be even.
Let FQR8 (fig. 48) be the course of any ray which starting
from P, after reflection at Q^ R and 8^ amvcs again at P,
and is there reflected in the direction PQ of the original pro-
pagation. Let F'Q'JR'S' be the course of another ray starting
from P' in a dii'ection F' Q' parallel to FQ', we have to shew
that after reflection at 5", this ray will proceed to P, and there
be reflected in the direction P' Q'. Join S'F'.
Then since the angle Q'F'A = QPA, and QPA=SFF, there-
fore the triangle PpP' is isosceles, and the ptTpciidiiiilar frmn /'
60 SOLUTIONS OF SENATE-HOUSE PROBLEMS, [1849.
on F Q = that from P' on PS. Similarly the perpendicular
from Q on QR = that from Q on Q'P' = that from P on P Q
since PQ is parallel to P' Q ; therefore the pei'pendicular from
Q on QR = that from P on PS. By similar reasoning it may
be shewn that the pei'pendicular from S' on PS = that from R'
on QR = that from Q' on ^i? since Q'R' is parallel to ^^ = that
from P on PS. Hence PS' is parallel to PS'^ therefore S'P' is
the direction in which R' S' will be reflected from S\ and P Q'
is that in which S'P will be reflected from P'.
The same proof may be extended to any even nmnber of
reflections. If the number of reflections were not even we might
still shew that P', S' were equidistant from PS^ but they would
be on opposite sides of it, as P', R' are of PP, and the pro-
positions would not be true in that case.
1849.
If the angle of a hollow cone, polished internally, be any
submultiple of 180°, a cylindrical pencil of rays incident parallel
to the axis will, after a certain number of reflections, be a
cylindrical pencil parallel to the axis, and of the same diameter
as the incident pencil.
Let fig. 49 represent a section of the cone and the light by a
plane through the axis CD of the cone, and let tn^m_^ be the
successive points when the ray PQ^Q^Q^... cuts the axis CD.
180°
(1). Let the angle ACB be an even submultiple of 180°=— —
90° ^''
suppose, or — .
Now the angle Q^m^D = ACD + CQ^m^ = ACD + A Q^P,
= 2ACD = ACB,
and the angle Q^^n.^D = BCD + CQ,^in^ = BCD + Q^Q^B,
= BCD + BCD + Q.m^D,
= 2ACB-
similarly Q^m^D = 3 A CB,
~ )
and Q„<»,D = uACB = 90°,
1850,] OPTICS. 61
or after the »"' reflection the ray will be pci*pendicular to the
axis CD^ and will proceed in a path exactly similar to that
already described, finally emerging in a direction parallel to
QyP^ and at the same distance as Q^P from the axis CD^ but
on the opposite side of it.
180°
(2). Let the angle A GB be an odd submultiple of 180°=
^ > ^ ^ 2n+l
suppose.
Now the angle Q.Q^B = BCD + Q,m,D^
= IACB -\- ACBhj the above,
and the angle Q.^Q^A = A CD + Q.^mJ)^
= \ACB + 2ACB,
= IACB;
similarly Q.^Q^B = IACB,
and Q,^Q,^,^A = ^-^ACB = dO'',
or after w reflection the ray will be perpendicular to the side
CA or CBj at which it has next to be reflected, and will there-
fore after that reflection return by the same path as it came by,
and will emerge in the direction Q^P.
Hence, whether ACB be an even or odd submultiple of 180°,
the emergent rays will form a cylinder equal in diameter to the
cylinder of incident rays, and having its axis coincident with the
axis of that cylinder, if the angle A CB be an odd submultiple
of 180°; or if the angle ACB be an even submultiple of 180",
the axes of the emergent and incident pencils will lie in the
same plane with the axis of the cone at equal distances on
opposite sides of it.
1850.
1. If a luminous point be seen after reflection at a plane
min'or by an eye in a given position, there is a certain space
within which the image of the point can never be situated, how-
ever the position of the plane mirror be changed ; find this
space.
62 SOLUTIONS OF SKXATK-Ilorsi: I'KOBl.EMS. [iSol.
It is easily seen that the distance from the eye of the image
foiTned by the mirror equals the actual length of the ray by
which the point is seen. This can never be less than the direct
distance of the point from the eye ; hence the image can never
be situated within the sphere which has the dii'ect distance
between the point and the eye for radius.
2. If a be the angle which every diameter of a circular
disc subtends at a luminous point, shew that the ratio of the
light Avhich falls on the disc to the whole light emitted is as
sin'^^a : 1.
About the Imninous point as centre describe a sphere with
radius unity : also with the luminous point for vertex and the
circular disc as base describe a right cone. Then the light
received on the circular disc : whole light emitted :: the portion
of the surface of the sphere Intercepted by the cone : whole
surface of the sphere.
Now by a known property of the sphere, the surface of any
portion of the sphere cut off by any plane is proportional to the
difference of the radius of the sphere and the distance of the
cutting plane fi'om the centre. Hence the surface intercepted
by the above cone : whole surface of the sphere :: 1 — cos|a
: 2 :: sin^'ja : 1, which is therefore the ratio of the light received
on the circular disc to the whole light emitted.
1851.
A sphere composed of two hemispheres of different refrac-
tive powers is placed in the path of a pencil of light in such
a maimer that the axis of the pencil is perpendicular to the
plane of jimctlon and passes through the centre : determine the'
geometrical focus of the refracted pencil.
Let r be the radius of the sphere, u the distance of the focus
of incident rays from the centre; v^v^v^ the distances of the
geometrical foci after the successive refractions, positive lines
being measured m the direction opposite to that of the incident
light ; fi^fx,, the refractive indices of the two hemispheres.
1851.] OPTICS. 63
Then i = - ^Vzi + ^. (1),
1 /i, 1
- =^ - (2 ,
(3) + (2)xl + (l)x^,,
i=-^.i^>,-i)+^,>,-i)}i+^:i,
which gives \\ the distance from the centre of the sphere of the
geometrical focus after refraction.
( <>4 )
ASTRONOMY.
1849.
1. Tlierc arc two walls of equal known height at right
angles to each other, and running in known directions ; shew
how to find the sun's altitude and azimuth by observing the
breadth of the shadows of the two walls at any given time.
And prove that the sum of the squares of the breadths of the
shadows will be the same whatever be the direction of the Avails.
Let a, h be the observed breadths of the shadows, h the
known height of the walls ; 6 the angle between the base of
the wall, the breadth of whose shadow is a, and the line joining
the shadow of the top of the line of intersection of the walls
with the bottom of that line, ^ the sun's altitude. Then
(y = tan 7- , and © = tan j-^ ^,-^, ,
i ' ^ (a^ + ly
are known. Let a be the angle between the wall whose breadth
is a and the plane of the meridian ; then a -- ^ is the angle
between the plane of the meridian and the vertical plane through
the sun, or the sun's azimuth. Hence both the altitude and
azimuth are known.
Also d^ +h^ = the square of the length of the shadow of the
Ime of intersection of the walls ; and the height of this line is
the same whatever be the direction of the walls, or a'' + // is
independent of that direction.
2. If the same two stars rise together at two places, the
places will have the same latitude. And if they rise together
at one place and set together at the other, the places will have
equal latitudes, but one north and the other south.
From the bisection of SS' the great circle passing through
the two stars S^ S' draw a quadrant of a great circle perpen-
dicular to SS' towards the north pole terminating in the
1850.] ASTRONOMY. (55
point r, and another towards the south pole terminating in T.
First, suppose the great circle containing these quadrants to
have its point which is nearest to P the north pole, to tlic left
of P (drawing the stars on the convex part of the sphere). Then,
in order that the stars /S', S' may rise together at any place,
its zenith must at some time in the 24 hours come to T\ its
co-latitude therefore must be TP. Hence if S^ 8' rise together
at two places, the co-latitude of each must be 2!P; hence the
latitudes of the places are the same. If the point of the circle
nearest to P lie to the right of P, the zenith of the place must
pass through T' in their daily path and therefore have the same
latitude, viz. 90' — T'P\ S., P' being the south pole.
If the same stars rise together at one place and set together
at another, the zenith of one must pass tlii'ough T and that
of the other through T m their daily paths; hence they will
still have equal latitudes, but one will be north and the other
south.
1850.
1. Prove that all stars which rise at the same Instant at
a place within certain limits of latitude, will, after a certain
interval, lie in a vertical great circle ; and detenuine those
limits.
This Avill happen when the zenith of the place comes to that
great circle of the heavens which at the time of the stars' rising
was the horizon of the place. Hence it can only happen for
those places for which the altitude of the pole is less than the
co-latitude : but the altitude of the pole is the latitude, hence
if / be the latitude, I must be less than 90° — / or / less than A,'/.
2. Shew how to find the days of the year on which the
light of the sun reflected by a given window which has a south
aspect will be thrown into some one of the lower windows of an
opposite range of buildings.
Corresponding to each window opposite, let that point of the
heavens be detennined, which lies in the same plane as that
window and the horizontal line through the reflecting window
pointing to the south, and at the same angidar distance from this
r
66 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1851.
line as the opposite window in question, and on tlie opposite
side of it. Let the north polar distance of this point or its
declination be then observed ; the reflected light will enter the
window in question on those days when the sun has this de-
clination. Suuilarly, the days when the reflected light will
enter the other windows may be detenniued.
1851.
Altitudes of the same heavenly body are observed from the
deck of a ship and from the top of the mast the height of which
from the deck is known : find the dip of the horizon and the
tnie altitude.
Let AB = x, BC= h (fig. 50) be the height of the deck
from the sea, and of the mast respectively ; OA = r the radius
of the earth. The difference (a) of the observed altitudes is the
angle^i^Oori>(9^.
Now cosCOE = 5 , and cosCOD = ,
r + h + x^ r + x'
.-. DOE = cos"^ 5 cos ^ = a,
r + li + X r + X
an equation for the deteinnination of x.
T
Then the dip of the horizon = OBD = sin~^ is known,
and subtracted from the altitude observed at B gives the true
altitude.
From the above equation we may determine x with sufficient
accuracy thus :
r + X \ {r + h + xY) r + h + x \ {r+x
\2r(h + x)}^- (2rx)^
or - — ^ -^^ — = sma,
r r '
omitting h and x in comparison of r ;
2(h + x) .2 « • /2a;\i 2x
-^ ' = sm^a - 2 sma — -\ ,
r \ r J r ^
f2x\^- , . h
I I = 4 sin « — —
h
, , — 9 sma coseca,
\r J ^ r '
or X = l^ sma — - coseca I
PART II.
F-2
( 69 )
PART 11.
EUCLID.
1848.
1. AB, CD, (fig. 51) are any two chords of a circle passing
through a fixed point O, EF any chord parallel to AB ; join
CE, DF meeting AB in the points G and H, and DE, CF
meeting AB in the points K and L : shew that the rectangle
OG.GH = OK.OL.
The triangles OCG, OHD have the common angle 0, and
^OCD = 180° - EFD = EFH,
= OHD,
since EF is parallel to AB ; hence the triangles OCG, OHD
are similar, therefore
OC:OG::OH:OD,
or OG.OH = OC.OD.
Again, L OCD = OEF = DKO,
since EF is parallel to AB ; hence the triangles OLC, ODK
are similar, therefore
OC:OL::OK:OD,
or OL.OK = OC.OD,
.-. OG.OH = OL.OK.
2. In a given circle inscribe a rectangle equal to a given
rectilineal figm'c.
Let AB (fig. 52) be a diameter of the given circle ABC.
Draw a square which shall be equal to the rectilineal figure
70 SULI'TIONS OF SEXATE-HOUSE PROBLEMS. [1848.
{Eu<\ II. 14,; through D any point of AB draw BE pei*pen-
dicular to AB, a third projxjrtional to AB and the side of the
square. Through E draw EC parallel to AB, meeting the circle
in C ; join AC, BC, and complete the parallelogram ACBF :
it shall be the rectangle required.
For C and F are each right angles, being the angles in a
semicircle, therefore ACBF is a rectangle. Also its area equals
AB.DE which equals, by constniction, the square which equals
the given rectilineal figure ; therefore it is the rectangle required.
3. Through a given point A (fig. 53) describe a circle which
shall touch a given circle BCD, and intersect another given
circle LEF in a chord passing through a given point G.
From G draw any line GEF Intersecting the circle LEF In
the points E, F ; join GA, and in GA produced if necessaiy, take
the point H, such that GA.GH = GE.GF. Through the points
A, H describe any circle cutting the circle BCD In the points
B, C ; join BC, and produce It to meet GA In K. From K draw
KD a tangent to the circle BCD. About the triangle AHD
describe a circle, it shall be the circle required.
And first It shall touch the circle BCD : for since KD
touches the circle BCD,
KD^ = KB.KC = KA.KH,
since one circle has been made to pass through A, H, C, and B ;
and therefore KD touches the cu'cle m question as well as the
circle BCD, therefore the two circles touch. Also the chord in
which it intersects the circle LEF will pass through G. For
suppose L to be one of the points In which It intersects the circle
LEF ; join GL and produce It to meet the two circles in M, M'.
Then
GL.GM = GE.GF = GA.GH, by construction;
also GL.GM' = GA.GH,
since H, A, L, M', lie in the circumference of the same circle,
therefore GM = GM' or the points M and M' coincide, and
GLM Is the chord in whicli the circles intersect, and the chord
passes through G as required.
1851.] EUCLID. 71
Hence tlic circle drawn as above described fulfils the required
conditions, and is therefore the circle sought.
1849.
Thi-ee circles are described, each of which touches one side of
a triangle ABC (fig. 54), and the other two sides produced. If
D be the point of contact of the side BC, E that of CxV, and F
that of AB, shew that AE = BD, BF = CE, and CD = AF.
Let AB, AC touch the circle GDH in G, H ; then
BG = BD, and CH = CD,
also AG = AH,
.•. AB + BD = AC + CD = semiperimeter of the triangle,
similarly, BA + AE = semiperimeter of the triangle :
.-. BA + AE = AB + BD,
and AE = BD ;
and similarly, BF = CE and CD = AF.
1851.
1. Let T (fig. 55) be a point without a circle, whose centre
is C ; from T draw two tangents TP, TQ ; also through T draw
any line meeting the circle in V, and PQ in B, and draw CS
perpendicular to TV; then SR.ST = SY\
Join CT, intersecting PQ in U at right angles ; draw CP;
it will be perpendicular to PT.
Since the triangles CTS, RTU are similar,
CT:TS::RT: TU,
.-. TS.TR = TC.TU,
or ST'^ - ST.SR = CT' - CT.CU,
= ST^ + CS^ - CT.CU,
.-. ST.SR = CT.CU -CS^:
but CPT, PUC are both right angles,
.-. CT.CU = CP'' = CV^
= SV^ + cs%
.-. ST.SR = SV^
72 [SOLUTION'S OF SEXATE-HOUSK PROBLEMS, [1851.
2. It" a ciix'lc be described round the poiut of intersectiou of
the diameters of a parallelogram as a centre, shew that the sum
of tlie squares of the lines drawn from any point in its cir-
cmnference to the four angular points of the parallelogram is
constant.
Join P any point in the circle with A, B, C, D (fig. 56) the
angular points of the parallelogram, and with O the centre of
the circle. Then, since OB = OD, the square of BP is greater
than the squares of OP and OB by the rectangle by which
the squares of OP and OD are greater than the square of DP
{Euc. IL 12, 13) ; hence
Bp2 ^ j)p. ^ Qg. ^ Qj)2 _^ gOP" is constant :
similarly, AP^ + CP-^ = AO'^ + OC' 4- 20P' is constant,
therefore also AP"' + BP'^ + CP' + DP"' is constant.
3. (a). Let B (fig. 57) be any point in the circumference of a
circle whose centre is A ; in AB take two points C and D, such
that ACAD = AB'; bisect DC in E, and draw EF at right
angles to AE ; in EF take any point G, then will the tangent
drawn from G to the circle be equal to GO.
Draw GF the tangent to the circle, join CG ; then
AG^ = GF^ + AF' = GF' + AB',
also AG' = GE^ + AE' = GE' + CE' + ACAD {Euc. ii. 6),
= CG^ + AB^ by construction ;
... GF' + AB^ = CG' + AB',
and GF = CG.
(/3). Describe a circle which shall pass through a given point,
touch a given straight line, and cut orthogonally a given circle.
Let D be the given point, BFL the given circle, KH the
given line intersecting AD in H.
In AD take AC : AB :: AB : AD, and HK a mean pro-
portional to HD and HC ; the circle through CD and K shall
be the required ciicle.
1851.] EUCLID. 73
For the centre of this circle will be at some point G of EG,
and since GF = GC, will pass through F, cutting the circle BFL
orthogonally at that point.
Also since HK is a mean proportional to HD, HC, or
HK"^ = HD.HC, HK will touch the circle through the points
C, D, K ; hence that circle fulfils all the required conditions and
is the circle sought.
( 74 )
ALGKBRA.
1848.
1. A walks to Trumpington and back by Granchester In
1^ hour, starting between 2 o'clock and 2f ; B walks the same
distance in the same direction in labour, starting between 2
and 2^ : find the chance that A overtakes B before he gets
home.
Unless A starts after B he cannot overtake him.
Now if he starts between 2 and 2^, it is an even chance
whether he starts first or not ; otherwise he cannot. Hence ^'s
chance of starting before B = ^.| = ^.
Again, A gets home between 3^ and 4, B between 3^ and 3f ;
therefore A has an even chance of getting home first ; therefore
also his chance of getting home last = ^, and chance of his not
overtaking B = chance of his starting first + chance of his
getting home last
= ^ + ^ = 1;
therefore chance of A overtaking -C = 1 — f ,
the required chance.
2. A paralleloplped is cut by three systems of parallel planes
given in number, parallel to the three pairs of opposite faces
respectively : find the total number of parallelopipeds formed in
every way.
Let in, w, j9, be the given number of intersecting planes
parallel to the three sides respectively; we thus have m + 2,
« -f 2, ^9 + 2 parallel planes in -three several directions.
Now, out of the first set of parallel planes we may make
^ '— — — - sets of two each. Similarly, out of the other two
sets we may make — ^-^ — — , — ~- sets respectively.
1848.] ALGEBRA. 75
Now each paralleloplped is formed by taking one out of each
of the above sets of two parallel planes, therefore the total
number of parallelepipeds will be
{m + 2){m+l) {n+2){n + l) (p + 2)(p+l)
2 ' 2 • 2 '
_ {m + 1) {n + l){p + 1) [m + 2) (n + 2) (^ + 2)
8 '
the required number.
3. If {^-a){i/-ma) = {n^-mo(.){x-a) (1),
and O' - a') (y - m/3') = {m^' - na) (cc - /3') (2) ,
shew that [—^-^-\x = -, ^-^r- ■
\aa pp J aa pp
Taking (1) {/3' — a) — (2) (/3 — a), so as to eliminate y, we get
m{^-oi){/3'-a'){^'-a)={{n/3-ma){l3'-0L)-{ml3'-na')[^-a)]x
- (n^ - ma) (^' - a') a + (?n/3' - na] [^ - a) ^8',
or m {(/3 - a]{l3"' - a 13') + [^' - a:)[oi' - ayS)} = [n- m)(/3/3' - aa>
-■n{(;S'-a>^+(;S-a)a'/3')l+m{(^'-a>'^+(/3-a)/3"0],
.-. {n-m){ (/3' - a') a/3 + (/3 - a) a^'] = (h - w) (/3/3' - aa') a-,
and dividing by aa'/S/3',
/J l_\ ^ g + g' _ /3-F /3'
"^Vgg' m'J da /3/3' '
4. (g). Shew that the integral parts of (3* + 1 )'""'' and
(3* + 1)'^'" + 1 are respectively divisible by 2'"'^^ where m is
any integer whatever.
The integral part of (S^+ir^^ is (3* + If'""^ - (3*- 1)"""S
since it is a whole umnber, and (3* — 1)^'"+^ is less than 1.
Now generally
^.«., _ y-.^. = [x-y) [[x^'+fl + try (a;-*- ^4/"'-0
+ a^/(a.— * + yn + ... + x'f (A),
and if £c = 3* + 1, y = 3* - 1,
x-y = 2, xy = 2,
.-. (3* + 1)"'""'' - (3* - 1)'^'"*' = 2 [{(34 + l)'-"" + (3* - 1)'""}
+ 2 {(3* + l)"^"-^ + (3i - I)"-""-'''} + . . . + 2"'].
76 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
This part of the question, then, reduces itself to shewing that
(3i+ 1)'-"" + (3*- l)'-"" is divisible by 2"'.
Again, the integral part of (34 + l)'''"' + l is {Si+1)''"'+{S^-1)'%
since this is a whole number and (3*— 1)^'" is less than 1 ; hence
the second part of the question reduces itself to shewing that
(34+1)'"' + (3i- 1)'^"* is divisible by 2"'^\ and therefore mcludes
the first part.
Now (3* + 1)*" + (3* - 1)"" = (4 + 2.34)'^" + (4 - 2.34f ",
= 2''"'{(2 + 34)'-'"+ (2-34n,
which is evidently divisible by 2^"^\ Also generally
_4«+2 I 4'i+a / 2 1 2\ f / 4» , 4»\ 2 2/ 4«-4 . 4n-4\
+ a.yK- +/"-),
-...+ (-)VY"};
.-. (34 + 1)*"+''' + (34 - 1)*'^'^ = {(34 + l)'-' + (34 - 1)'}
[{(34 + 1)*" + (34- 1)*"} - 2'-^{(34 + l)*«-^+ (34-1)^"-^} +...+ {-Y2'"l
which, since (34 + 1)'" + (34 - 1)*" is divisible by 2'"^', is divisible
by 2'"+", and therefore a fortiori hy 2'"'"+' or 2^''''*'^*\.
Hence, whether ?n be of the fonn 2n or 2n + 1,
(34+1^"+ (34 -If"
is divisible by 2'"^*,. and both parts of the proposition are true.
(/5). Prove that for a given integral value of a, there are
(1), a integral values of b which will make the integral part
of (a + b'-Y'"^' divisible by 2'"+^ ;
(2), a integral values of b which will make the integral part.
of {a + b'~Y"'^' + 1 divisible by 2'"^^ ;
(3), 2a integral values of b which will make the integral part
of {a + UY" + 1 divisible by 2'"+\
(1). The integral part of (^4+ af'*' = (&4+ af"^' - [U-af'^-'j
provided b^ — a < I and > 0, i.e. if b lie between a^ and [a + 1)"^,
which gives only 2« values of b.
1849.] ALGEBRA. 77
Now by equation (A),
+ [h - «^) [{b'- + ay-'-' + {hi - aY"'-'} +...+ (/._ aTl
Now since b is only to have (a) values, we may make b — (i^
even, and then the problem is reduced to shewing that
{bi + ay"+{bi-af"
is divisible by 2", which will h fortiori be tme if
{b'- + ar + {bi-aY%
{the integral part of {b^' + af' + 1} be divisible by 2"^', and
therefore (1) reduces itself to (3).
(2). The integral part of {a+b^-f"'^'+l={a+b^-Y"'^'+{a-biy"'^\
provided a — b^ < 1 and > 0 ; therefore b must lie between n^
and [a— 1)'\ and it will appear by a precisely similar process to
the above that (2) reduces itself to (3).
(3). The integral part of (a + b^Y'" + 1 = {a+b'-f"' + {a - 7>i)*",
provided a — b^< 1 and > — 1, i.e. if b lie between [a— 1)'^ and
{a+ ly, giving only (4«) admissible values of b.
And if we flirther make b — d^ or d^ — h even, the number
of values of b will be reduced to (2a) .
Then, these conditions being satisfied, (3) can be proved as in
(a), only writing 5* for 3* and a for 1.
Hence the propositions enimciated are tnie.
1849.
1. A quantity of com is to be divided amongst n persons,
and is calculated to last a certain time if each of them receive
a peck every week ; during the distribution, it is found that one
person dies every week and then the coni lasts twice as long as
was expected : find the quantity of com and the time that it lasts.
Let X = the number of pecks of corn,
y = weeks it is expected to last,
then - = the whole number of persons = n ;
y
or X = ny (!)•
78 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1H49.
Also n = number of pecks distributed in tlie 1st week,
V — 1 = 2nd
n - r + I = J-""
and it lasts 2i/ weeks ; therefore
n + {n — I) + {n — 2) + ... + {n — 2?/ + 1) = whole quantity of corn,
or {{2{n-i/) + l}?/ = x (2);
.-. by (1) 2{n-y) + I = n,
n + I
_ 7i[n + \)
^ - 2 '
and 2y = n ■\- 1,
which determines the quantity of corn, and the time it lasts.
2. If ,^, C,. be the number of combinations of m things taken ?•
together, and jy be less than m and ??, shew that
We have (1 + xY{\ + .-r)" = (1 + a-)'"^".
Now the coefficient of £c" in (1 + .t)" is vOfx'. also the coef-
ficient of x^ in (1 + a;)"'"^" must be the sum of the jiroducts of the
coefficients of a;' in (1 + .x)'" multiplied into the coefficient of
af"" in (1 + x)" taken for all values of r from 0 to j>. Hence
3. If ^Gr be the number of combinations of n things taken
r together, prove that if a be an integer greater than 1, then
will „,(7,„ be greater than („C,.)".
1849.] ALOEBILV. 79
The total number of combinations which can be made out
of ?w things is „„C,.„. But if we divide the na things into a sets
of n each, and restrict ourselves to those combinations containing
ra things which can be made by takuig r out of each set, we
shall get, since ,, (7, combinations of r things may be made out of
each set of w things, [„C,.)" combinations.
Hence, since (7.„ includes every possible mode of fonnationa,
and („6'^)" only one particular one, it is clear that
4. If ^ be greater than unity,
1 1 1.2 1.2.3
and if it be less,
1 1 1.2 L^__ «
1-j, 1 +i? (1 +p) (1 + 2p)^ ^ (1 +p) (1 + 22j) (1 + 32)) ■
We have in general, when^ is greater than unity,
1 \__ _ M + 1 1
p — \ ^; + w 2^ -\r n' 2^ — X"*
1 1 ?? + 1 1
J) — ^ J9 + 7? ^? + 7i '/> — 1 '
therefore, putting successively n — 1, 2, 3 ,
112 1
+
jj — 1 2.) ■\- \ ^ + 1'^> — l'
1 +^f^+
^j + 1 ^^ + 1 V7? + 2 y> + 2 ^ - 1
1 1.2 1.2.3
+ T— ^T7 n^N +
■^ + 1 "^ (^+l)(^, + 2) ^ (^>+l)(7.-+2) •;>-!'
1 1.2 1.2.3
+ 7 — r-^, — r^x + 7 — , ,x/ , ■ nN/.. ■ ox +••
2>^\ (i?+l)(i^ + 2) " (^j+ l)(iJ + 2)(^^ + 3)
80 SOLUTIONS OF SENATE-HOUSE PKOBLKMS. [IHot*.
Kj) be less than unity,
_^ I _ ^ {n+_l)p _i_ .
1 —J) 1 + np 1 + lip ' 1 — /' '
therefore, putting ?« = 1, 2, 3,... successively,
I _ 1 2p 1
1 -i> ~ 1+p l+p ' \-p '
- ^ 2p / 1 3/> 1 \
~ 1 +/?''" 1 +p li + 2^ "'' r+ 2p • 1 -pj '
~ l+p'^ [l+p)[l^2p)^' "^ (l+^)(l + 22^)(l + 3/^)^'
-f
5. A bag contains three bank-notes, and it Is known that
each of them is either a £5, £10, or £20 note ; at three suc-
cessive dips into the bag (replacing the note after each dip) a
£5 note was drawn : what is the probable value of the contents
of the bag ?
There are six possible states of the bag, viz.
(1). 3 £5 notes in which case the value would be £15.
(2). 2 £5 and 1 £10 £20.
(3). 2 £5 and 1 £20 £30.
(4). 1 £5 and 2 £10 £25.
(5). 1 £5 and 2 £20. £45.
(6). 1 £5, 1 £10, and 1 £20 £35,
and these are all a priori equally possible.
Now the chance of the observed event in case (1) is 1,
(2) or (3) (ror|,
(4) (5) or (6) (i^or^,
therefore the probable value of the contents is
1 X £15 + A (£20 + £30) + sV(£25 + £15 + £35)
1 + 2 X I, + 3 X i
— * ~ 4fi 1
27
_ 4?nio
— ^mi
= £19 15.?. V^^d.
ALGEBRA. 81
1850.
1. Prove that the sum of the fractions which are intermediate
in magnitude to any two nmnbers vi and n, and liave 3 for a
denominator, is n' — ni^.
The fractions, together with the intermediate whole numbers,
will be
3w + 1 ^m + 2 3n - 1
3^"' 3 ' 3
. /3m + 1 3« - 1\ (3^2-1) - 3»«
whose sum is I — 1 — — 1 ,
_ [m-\-n) [^[n-m) - 1}
_ _ ^
and the sum of the intermediate whole numbers is
(m+1) + (?» + 2) +...+ («- 1),
» — 1 — m
= [[m + l) + [n-l)^ ,
(■w + r?) {ii — m— 1)
^ 2 '
therefore the sum of the fractions is
(w + n) (3 [n — m) — 1} {m -f n) [n — m — 1)
2 2 '
= n' - m\
2. There are a number of comiters in a bag, of which one is
marked 1, two marked 2, up to r marked r ; a person draws
a coimter at random, for which he is to receive as many shillings
as the nmuber marked on it : find the value of his expectation.
Since the person is as likely to draw one counter as another,
the value of his expectation
total value of contents of bag
number of counters in bag '
1'^ + 2^ + ...+ ,.«
= l + 2+...+ >- '^'^^'''^''
82 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
Now I' + 2' + ... + r' = „ ,
o'{r +l)(2r+l)
6
r(r + l)
1 +2 +... + r= \^ ' \
r(r + l)(2r + l) r(r + l)
hence his expectation = -=^ '-^ 1 r — -
= — - — shillings.
3. If a, J, c be in harmonic progression, shew that
111 1 ^
- + - + 7 + 7 = 0.
a c a — b c — o
T * 7> 1 1 1
>S' /3 + a' /3-a'
then - 4- - = 2/3,
a c
and a - b = -p: -^ ,
_ a
-"/3(/3 + a)'
7. 1 1
_ a
"/3(/3-a)'
^._ _J_ _^ _J_ ^ /3(^-a) _ /3(/3 + a)
' ' a — b c — b a a '
= - 2/3,
1111..
.-. - + -+ 7 + 7=0.
a c a — b c — b
4. If there be z counters of which z are marked m\ z .n,...
with or without other marks; 2!„,^, w, w, with or without other
marks; 2;^,,„ ,,,y... marked w, w, ^j, ^...; the number mimarked
is s — 22 „+ 22;,„,„— 22;„,,„,p+..., 2 involvmg all combinations.
1850.] ALGEBRA. 83
Let Z>,,, denote the operation of selecting from the counters
those marked with ??«, D^ those marked with w, &c. Then it is
manifestly the same thing whether we first select from the heap
those marked ;/?, and then from these, those also marked n ; or
whether we first select those marked w, then from these, those
marked m ; or at once select those marked 7n, n. Tliis may he
symbolically expressed thus :
D D = D D = D ;
similarly, we have in general
D D D ... = D D D ... = ... = D ...: (1).
Also 1 — Z),,, will denote selecting those unmarked with m^
1 — D^^ those immarked with ??, &c. Hence the whole number
unmarked _ ( i _ 2) j (i _ i)J . . ,z^
= (1-22) +2i) -2i>,, „+...)z',
since the spnbols D^^^^ A.v have been shewn (equation (1)) to
be commutative
■^ m ' ^ m,n ^ m,n,p ^•••1
the required number,
5. Prove that
a;' + / + (a; + y)« = 2 (.x" + xy^- ?/)* + SxY [x + yf (ar* + xy + f] ;
and if x^ + xy -{• y^ = a, xy {x + \j) = i, and n be any positive
integer, shew that
^^»+y^-+(^+^)^»^2a" + n(n-2)a-6- + "("-'^^;;;'^^"-'^a-^-
/^(»-r-l)(7^-r-2)...(ri-3r + l) ,_„.^,.
3.4. ..2r
(a). Let 2 be a quantity, such that
X + y + z = 0',
then x"" + f + [x^-yT = x'" + /" + z'\
g2
84 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
and x'' + xy + y^ = {x-\- yj — xy^
= -z{x + y) -xy,
= - {yz + zx + xy)j
xy{x + 7/) = -xyz-
therefore, taking the notation of the latter part of the question,
yz + zx + xy = — a,
xyz = — by
therefore x, ?/, s, are the roots of the equation
.•. {^ — x){^ — y){^ — z) = ^^ — af + hj identically j
-i)(-D(-i)--M'
l-|.(«-jy,
.-. log(l-|)+log(l-|)+log(l-|)=logjl-^,(a-|
x + t/ + z , 1 a;' + / + ^' , , 1 x" + /" + z'" ,
•• f +2 f ''"••■"'"2m f" "'"•
1 / 6\ 1 / i\^' 1 / i\"
= rr-|; + 2rr-|j +-+s|-^«-|j+-;
therefore equating coefficients of ^j ,
i_ (aj--"- + y2» + 2'-*«) = 1 a" + -^ {n-l){n-2) ^„^,^,
2n n n—1 1.2
J_ (n-2)(n-3)(n-4)(7^-5) ,
■^71-2 1.2.3.4 "" ^ +•••
1 [n-r){n-r-l)...{7i-?,r+\)
»i — r 1.2. ..2r
n 2 2.3.4
(9i-r-l)...(n-3r+l)
^ 2.3. ..2« "* ^ +•
1850.] ALGEBRA. 85
.-. x'" + /" + [x + T/y = 2a" + n{n- 2) a"-'b'
{n-S){n-A){n-5) ^,.^^, _^^
n
+ - 3.4
n{n-r-i)...{7i-Sr+l) ^
^ 3.4.. .2r "" ^ ■^••••
Hence putting n = 4, we get
a;' + y + (a; + yY = 2 [x" + xij + ff + S{x'' + X7/ + f) xY {x + yf.
/aV'^''
6. (a). If a be less than h, prove that ij-j is increased by
adding the same quantity to a and b.
(/3). And if w be greater than 1, shew that
by means of this fonnula prove that ^
(a^ + a, +. . .+ a J" > n\a^. . .a^.
(a). Since a is less than &, we may put a = h — c, where c is
a positive quantity less than b ; then a and 5 will be increased
by the same quantity if we increase Z>, c remaining constant j
^y "V~^j = a: suppose;
•. loga;= {2b-c)\og(^l -^ ,
/ c\ /, 1 c 1 c'
2 l\c /2 l\c' /2 1
ti-i
which is manifestly negative, since the coefficients being of the
form (
- j are positive ; and the absolute magnitude of the
series is diminished by increasing b if c remain unaltered ; hence
( V 0+6
r) , is increased by adding the game
quantity to n and h.
86 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
(/3).* First, let J be greater than unity = 1 + - suppose;
we have then to shew that
nj L n—1 J V " — 1 ,
these quantities when expanded by the Binomial Theorem become
, + ^ + __^' + —i *+•■■'
, n — \ ., V n—\j\ M— 1/ 3 .
and 1 + a; + — 7-T x- + -—- x +... ;
all the terms of both series are positive, and each term of the
first series greater than the corresponding term of the second, or
1 + 7J "(^ + ^1) '^^^ U) >toj '
when a is greater than h.
Cb . X
Next, suppose j less than unity = 1 — - suppose, we have
to shew that
l-i^^ > J % or 1 ="
w/ I w — 1 J V n — 1
(. -ii-i
1 j > (1 — x)''":
these quantities when expanded by the binomial theorem become
71 — \ „ V 71— l) \ 71— \,
1 + ^^+ 1.2 ■'" + 1:2:3 ^ +■■•'
7^ .y \ 71 \ 71/ .,
and 1 + X + ^^ a.^ + ^-^^^ ^^ + . . . ;
* 'This part of the solution is given by Mr. Thacker in a recent number of
the Vamhridfie and Dublin Mathematical Journal, No. xxv. p. 8L
1850.] ALGEBRA. 87
all the terms of both series are positive, and each term of the
first series greater than the con'esponding term of the second,
and therefore the proposition is true in this case also.
Now let the n quantities «j, a^v^n? ^® ^^ ascending order of
magnitude ; then
V n / ' V na^ ) '
^"^ 1'+ [n-l)a, I '
by the above,
^rt. -I- «,+...+ a
n-l
> a
similarly ( -^ 5__^ « j > a_^ f _! * "
w-l ' '
n-i f ^ I ^ , I ^ \ n-2
?
>
n-l ' H
, > a ,a :
hence by multiplication
or [a^-\-a^^-..,+ aJ' > n{a^a^..,aj.
7. If - be the r'^ fraction converging to — , and n' be the r*^
remainder in the process for finding the successive quotients,
prove that
m p _ p
n q pq '
Let ^ , ^ be the {r - 2)'^ and (r - 1)* converging frac-
9'r-2 2'r-l
tions respectively; m the r'^ quotient, n' the (r— !)'•» remainder;
then i^ = »'^>r-, +i^r-25
88 SOLUTIUNS OF SENATE-HOUSE PKUBLEMS. [1850.
And the fx"actioii — may be derived from - by wTitiug rii-\- —r,
for 111 In the above expressions for p and q ;
in
'«' + :^' ) i^r-i -^Vr^
Now — is in its lowest tenns, and n is prime to n",
.-. m = [m'n" ■^n)p^^^ + n"p^_^^
n — [m'n" + n) q^._^ + n'q^_,^^
, m p mq ^ np
anci '^ -^ .
n q nq
Jig- *
8. Find the probability of drawing a black and a white ball
the same number of times from a bag which contains an equal
number of each ; the balls being dra^\ai one by one and replaced
after each draAving, and the nmnber of drawings being the same
as the number of balls in the bag, but this number is unknown,
any number from 2 to 2n being equally probable.
Suppose there are 2x balls in the bag; the number of
drawings will then be 2a;, and the number of possible ways in
which the balls may come out = 2'''''".
Of these the number of favourable cases equals the number
of permutations of 2x things taken all together, of which x are
of one kind and x of another,
_ 1.2. ..2a;
~(l.2...a;y^'
_ (a; +1) (a; + 2)... 2a;
~ 1.2. ..a; '
therefore chance of proposed event on this supposition
_ J_ (a- +1) (a; + 2)... 2a;
~ ^r 1.2. ...r
I851.J ALGEBRA. 89
Now the chance of there being 2x balls in the bag = - what-
n
ever number > 0, > «, x may be. Hence the chance of tlie
proposed event
^ ^^ !^l!i . .1 in+l)in + 2)...2n
n (2M 2M.2 2'"' 1.2
1851.
1. If y , y7 be fractions in their least terms, the denominators
of which do not exceed a given nmnber ??, the fonner fraction
being given, and the latter detennined from it by taking for
a and b' the greatest values of x and y (?/ not greater than w)
which satisfy the equation hx — ay = 1, then of all the fi'action
in their least tenns, the denominators of which do not exceed w,
the fraction j-, exceeds y by the smallest quantity.
,,,. , a a ha — ah' 1
Wehave _ _ _ = _^^ = _ ,
since a', h' are values of x and y in the equation hx — ay = 1.
Let ^ be any other fraction in its least tenns whose de-
nominator does not exceed /i, then
a a ha — a/3
^~h^ ~ir^ ~ h^
m being some integer greater than 1 ; then a and ^ are values
of X and y in the equation
hx — ay = m.
Now this equation is satisfied by
X = ma ± qa^
y = mh' ± qh^
where q is any integer.
Again, the successive values of x and y which satisfy the
ccpiation ;,,y _ ^,y ^ 1^
= Ta say,
90 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
differ by a and b respectively ; hence, as b' is the greatest value
of y, less than n, in this equation, b' + b> n, therefore h fortiori
mb' + b> n. Hence /S cannot be of the form mb' + qb^ since
it is less than n.
Neither can it be of the fonn mb' ; for then a would = ?««',
and p would not be in its least tcnns.
Hence yS must be of the form mb' — qb :
.•. /36 = [mb' — qb)b < mbb'j
a
a m 1
13
b ^ mhb' ^ bb' '
a a
or of all the fractions in then* least terms, whose denominators
t
do not exceed ??, y? exceeds r by the smallest quantity.
2. The sum of the series -^ + -^^ + -^^^ + ... to infinity,
where S is positive, is greater than (2^ — 2) "^ [^^ ~ ^)i ^^^ 1®^^
than 2* -=-(2^- 1).
Let 8 be the sum of the given series.
Then
1 1 /_2_ ^ _8_
> p+3 + 2 V2^* "^ 41+a "*" 8^3 "*■
^ P "^ 2 V2^ ^ 4^ "^ 8"^ "^
>lH-i '
2 2^-1
2«- 1
1851.] ALGEBRA. 91
Again,
1 /J_ J_\ /J_ _1_ _1_ Jl_\
1 1 1
^ jd ^ 2'» 4*
1
<
2«-l
2^ — 1 2^
Hence S lies between -i — ~- and
2* - 1 2^ - 1 *
3. Solve the equation
1 1 — X I I - a ^
X + a 1 = 0 ;
1 — X X 1 — a a
and thence infer the resolution of the first side of the equation
into factors.
We see at once that the given equation is satisfied by a; = a.
Again, for a write ; , then
1 _ 1 _ _ 1 -a
1 - a ~ 1 ~ ~~ar *
I — a 1 — a ,
= • = - a.
a 1
1 -a
Hence the given equation becomes
1 1 - X , 1 \ - a ^
X + a - 7 + — — = 0,
I — X X 1 — a a
of which a or is a root.
92 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Similarly, ; — is a root of the given equation.
But a_-l a 1
a - \ 1 - a '
a
therefore the roots of the given equation are a, , — ;
and the first side may therefore be put into the form
x[l —x)
4. From the equation
ah[c + d-e -/) + cd{e +/- a-h) + ef[a + h-c-d) = 0,
detennine, in terms of &, c, (7, e,y, the ratios a — c : a — d, and
a — e : a—f\ and shew that the relation between the six letters
may be expressed in the form P = Q^ where P and Q are each
of them the product of three differences of pairs of letters, or
in the form R= 8^ where R and S are each of them the product
of four differences of pairs of letters.
(a) From the given equation we get
_ cd[h - e -/) + ef{c + d-b) ^
^~ b{c + d-e-f) - cd + ef ^
_ c[b-d) (e+f) - he' + ef{d-b) + 6'd
•'• "" """ b[c+'d-e-f) -cd + ef
^ "^1 b{c + d-e-f)-cd^€f
=.-(b-d] ('-')i'-f) (1)
^ ' b{c + d-e-f) -cd+ef ^^'
Snmlarly, a-d=-{h-e) ^^ f^^^^_^_^ J ^^ ^y 5
— c _b — d c — e c — f
- d b — c' d— e' d — f
(2).
1851.] ALGEBRA. 53
In the same manner it may be shewn that
a ~ e _h - f e — c e — d ,,
^irj-h^e'T^c'J^d ^^^'
(yS) From (1) we see, by interchangiug c with e, d with /,
which does not alter the original equation, that
^ ^ ' l>[e-\-f-c-d) -i- ef- cd^
a — c _ h — d {c — e){c —f) ^
' ' a — e ^ ~ f (e — c) (e — c?) '
••• («-«) (*-/) {e-d) = [h-d) {c-f) {a-e),
which expresses the relation between the six letters in the fonn
P=Q.
(7) Again, by (2) we get
(«-c) [b-c) {d-e) (d-f) = {a-d) [h-d] (c-e) {c-f),
which is in the form B = S.
5. If d^ + 1 be exactly divisible by j), and — be converted
into a continued fraction, until two consecutive reduced fractions,
— , — , , are found, such that p^ > n < n. then
n ^ n ' ^ '
j^ = [na — mjpf + w".
It is manifest that
{na-mp) =np [-- -)
., ., (m mV
by the property of Continued Fractions.
-r, m m 1
But — = + — -, ;
n a nn
.'. (na — mpY < ^, ;
and n''^ > p ;
94 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
therefore, h fortiori^ [na — mpf < p^
and n^ < p ;
.*. {na — inj})'^ + n^ < 2_i).
But {na — mj)f + n^ = n\d^ + 1) — 27nna.p + m^p\
which must be divisible by p, since a^ +1 is so :
and {na — mpf + ri' has been shewn to be < 2^, and it must
be positive ; therefore we must have
{na — mpY + n^ — p.
6. If {a+yS+7+...}^ denote the expansion of (a+/3+7+...)",
retaining tho^e terms iVa"/3V^''-" only in which
& + c + ^+ ... l!(>p— 1, c + d+ ... ■;^ p — 2^ &c. &c.,
then
*This theorem may be put into a rather more convenient
fonn by wi'iting x — a. for a- ; we have then to shew that
{x - a.y = x''-n {ay {x + /S)"'^ + ^ll^zil (« + ^Y {x + ^ + 7)""''
or, writing a^ for a, a,^ for /3, &c.,
(a. - «,)• = a;" - « (a.)' (* + «J- + '-^^^ (a, + «,f (.r + a, + ocj""'
- "'""lilT"'' («.+«^+='.l' (-+«.+«.+=•.)"-' + (1). '
The proof of this depends on the expansion of the quantity
{a, +«,+ ... + «,}".
* For tlie solution of this problem we are indebted to Mr. Cayley.
1851.] ALGEBRA. 96
Expanding by the Binomial Theorem,
(a^ + a,+ ...+a,)''
To pass to {a, + a^ + . . . + «/,}''. The sum of the indices of
Og, ttg, ... a^,, are not to exceed^ — 1 ..., and generally the sum
of the indices of a,., a;.^^, ... a^,, are not to exceed p — r + 1.
Hence in (a.^ + ag + ... + a^)*", the required conditions will be
satisfied, if only the sums of the indices of a^_,.+2, ap_,.+3j ■ • • «p do
not exceed *• — 1, and the sum of the indices of a^.^^,, a^,^^^, ... a^
do not exceed r — 2, and so on. And this will be the case
if, considering a^ + ag + . . . + a^_^+j «^ one quantity^ we replace
(a^ + a3+ ...+a^)'by
K^a + «3 + • • • + Vm) + V-+2 + • • • + a;,}^
Hence {a, + a,+ ... +a,|^ = a,^ + |a/"M(«2 + "3+ ••• V^+i^f)!'
+ f ^^ «/"' {(a. + "3 + • • • + a,_,.) + a^_^Y + ...
+ pa^{a, + a^ + ... 4 a,}""' (2),
the last term (a., + a^+ ...+ OLp)" being of course rejected alto-
gether, since the sum of the indices exceeds^ — 1.
Now the coefficient of - — - — ... (- a,)*", on the
12 r
left-hand side of equation (1), is cc"~'.
On the right-hand side, it is, expanding each of the quan-
tities in the brackets ( } by (2),
(aj + a, + ... + a,J"-'--^((a,+ ... + a,J]'(a^ + «, + ... + a,,,)"-'-^
n-r-1
r+2/
96 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [I8iil.
or, if wc write y fur x + a.^+ ... + a,^,, to for a.^ + ... + a,,^,, the
coefficient on one side is (^ — w)""'', on the other
«-r " ' f 11/ , \M-'-l
» — r » — y — 1 , ,,, , Ml-,-!!
+ —J 2 ^"^ "^ "''■^•'^ ^^ "^ ^'^' ^ °''*='^ '
+
Therefore we have to shew that
n — r n — 7' — 1 , lo / , , \»i-)-->
+ _ _ {a, + a ,,r {y + a .,, + a ,3)" "-...
the theorem itself, wiiting n — r for n.
The coefficients of a^**, on each side of the equation (1), are
obviously equal. If then the theorem hold for the indices
1, 2...(m — 1), it is proved to hold for the index n. But it
obviously holds for the index 1 ; therefore it holds miiversally.
7. If = a + — ^-— , and — —, — = a + , , :
mp + X ]} + It mj) + X p + 7^ '
then supposing ic and u to vanish when x vanishes,
1 1 . ,s
-7 — = m (a. ~ a. .
u ti
Since u and u' vanish when x vanishes, we have
1 B 1 . )8
— = a+— , — , = a + — ,
mp J) tnp p
Hence, by the first of the given equations,
p + u 1
Tnp + X m
X
.'. p + u = au {mp + a?) + ^ H
oc
= w ( 1 - mB + oLx) + » — ;
1851.] ALGEBRA. 97
.'. u (m8 — ax) = — ,
m
m'/3 1
.'. ma = - .
X u
Similarly, by the second of the given equations,
m^^ , 1
ma = —:
X u
the required result.
11,
-, = m a — a
u u
H
( ^'H )
PLANE TRIGONOMETRY.
1848.
A cannon-ball is moving in a direction making an acute
angle 6 with a line drawn from the ball to an observer; if
V be the velocity of somid, and nV that of the ball, prove
that the whizzing of the ball at the different points of its
com'se will be heard in the order in which it is produced, or
in the reverse order, according as /? < > sec 6.
The whizzing will be heard in the ordei' in which it is pro-
duced, or in the reverse order, according as the sound or the
ball moves more quickly towards the observer. Now the velo-
city with which the ball moves towards the observer = wFcos d ;
hence the whizzing will be heard in the natural or reversed
order, according as
V >< n V cos 0,
or no sec 0,
the required condition.
„ j^ sin (a — jS) _ sin (a + ^)
sin yS sin ^ '
shew that cot /3 — cot 0 = cot {ol + 0) + cot (a — /3).
Since
sin (a — /3) sin (a + 0)
sin /3 sin 0
1 1
sin /3 sin (a + 0) sin 0 sin (a — yS) '
sin {a + 0-^) _ sin (g - /3 + 6') ^
sin yS sin (a + 0) sin 0 sin (a — /3) '
.'. cot yS - cot {a + 0) = cot 0 + cot (a - /3),
or cot 13 — cot 0 = cot {a + 0) + cot (a — /S),
the required result.
PLANE TRIGONOMETRY. 99
1849.
1. If COS a = cos ^ cos <^ = cos yS' cos 0',
and sin a = 2 sin ^(/> . sin ^0',
shew that tan \a = tan \j3 tan ^^8'.
Since sin a = 2 sin \^ sin |^<^',
.-. sin' a = (2 sin' :|<^) (2 sin' ^ <^')
= (1 — cos 0) (1 — cos 0') ;
, / cosa\ / cos a
,1 — cos a = 1 :x 1 -
cos/Q/ V cos/37 '
.'. cos a = sec /3 + sec /S' — cos a sec /8 sec /8',
, sec B + sec yS' cos B + cos /3'
and cos a = :, ^^ 7^ = 7^ p=; ;
1 + sec p sec p I + cos p cos ya '
2 <* _ 1 ~ cos a _ 1 — cos y8 — cos /8' + cos /3 cos y8'
2 1 + cos a 1 + cos /8 + cos /8' + cos /8 cos yS'
^(l-cos^)(l-cos^;) g ff
(l + cosy8) (l+cosyS) 2 2 '
a /8 yS'
and tan - = tan — tan — .
2 2 2
2. Find •=• from the equations
[a + h) sin d + [a - h) cos 6 = {ci' + Jr)\
a sin' e + b cos' 0 = {Zah)K
Squaring the first of the given equations, we get
{(i'+V') [mi'd^- cos'^) - 2ah [cos'S-sm'S) + 2(a'- J') sin^cos^=a'+ b';
.-. (a' - &') sin 2^ - 2rtZ> cos 2$ = 0.
Let - = tan </>, then this equation gives
sin 2 (^ - 4>) = 0,
whence 2 (^ — ^) = 0 or tt ;
.". 6 = (f) or (f) + ^TT.
11 -J
100 SOLUTIONS OF SENATE-HOUSE PROBLEMS, [1849.
Taking ^ = </>, the second of the given equations gives
a siii^ <^ + i cos' 0 = (3aJ)*;
'fTI + ^7-^ = ('«^^^-'
a J \ a
• • a = (3«&)*,
and 72-3y+l=0;
o o
« 3 1 , , ,
••■5 = 2±2(-')' (')•
Again, taking ^ = ^+^7r, the second of the given equations gives
a cos^ (f> — b sin'* <f) = (3aZ>)*,
a — b ,„ ,, ,
1 = {^^^)H
{a'-^bj
.-. d'- W= {a' + bf {3abf,
a* - Id'V + b^ = Sab{d' + V'),
(a^ + yy _ ^.d'b'' - dab (a' + J'^) = 0 ;
... (a^ + J2 _ 4^J) ^^2 _^ J2 ^ ^j^ ^ 0 .
therefore, fii-st, a''* + J^ — 4a5 = 0,
^'Z Kb.
and ^ = 2±(3)i (2):
1850.]
PLANE TRIGONOMETRY.
or, secondly,
d' + h' -V ah = 0,
and
••• ?=-i±i(-3)*
101
(3).
The six values given by (1), (2), and (3), are all the values
which Y admits of.
o
1850.
1. If through the angles of a square four straight lines be
drawn externally, making the same angle a with the successive
sides, so as to form another square, find its area.
Let a be the side of the interior square : then the area of the
exterior square = interior square + four triangles each equal to
^d^ sin a cos a, or = d^ + 4 x ^a^ sin 2a = d\l + sin 2a).
2. Shew that
2tan-Mtann45°-a) tan 1/3} = cos"' /J^^1^L±£^^
^^ ^ / 2A-J Vl + tanacosyS; '
if a be < 45°.
In general cos (2 tan"' j:;) = 2 cos^(tan~'cc) — 1
2
1
Let X = tan* (45° — a) tan^/3, then
^ ^ 1 + tan(45 -a)tan'^/3
1 — tana 1 — cosyS
_ 1 + tang 1 + cos^
1 — tana 1 - cos/3
1 + tana I + cos/3
_ tana + cos/3
~ 1 + tana c<>s/3 '
102 SOLl'TK.lNS OF SHNATE-HuUlSE rKOliLK.MS. [I«o0.
.-. 2 tan-' {tan*(45" - a) tan^/3} = cos- ( tan« + cos^\
If a were > 45°, the given expressions would become imaginary.
3. Draw AB and AC (fig. 58) at right angles to one
another, and make AB equal to twice AC] produce CA to D
until CD is equal to CB: prove that BB will be the side of
a regular pentagon inscribed in a cii'cle, of which AB is the
radius.
Also, if with centre B and radius BA we describe a circle
AEF^ of which ABF is a diameter, and make AE equal to AB^
then FE will be the side of a regular pentagon circumscribing
a circle, of which A C is the radius.
(a) Let AC = a, then AB = 2a ;
.-. CB=CB= [AB' + A Cy = bhi ;
.-. AD= (54-1) a,
BD = {AD^ + ABy
= (6 - 2.5* + 4)*a
= (10 -2.5*)* a
, (10-2.5*)*
= 4a ^^ —
4
= 2 sini7r.2a
= 2 sin lir.AD ;
therefore i?i) is the side of a regular pentagon inscribed in
a circle, of which AD is the radius.
{^) Again, FE'' = AF'-AE'
= {2.ADY - AB'
= 4(6-2.54)rt"'' - 4a'
= (20 - 8.5*) a'
= 2^(5-2.5*) [aY
= {2tsinl7ryAC';
therefore FE is the side of a regular pentagon circumscribed
about a circle, of which AC is the radius.
1850,] PLANE TRIGONOMETRY. 103
4. If (?, hj c, be the sides of the triangle ABC, 2^1 ?> ^ lines
bisecting the angles drawn to the opposite sides, and p\ g-', r
these lines produced to meet the circle which circumscribes
the triangle; shew that
COsi^ COsii? COsiC 111
— -I- = — -k^ = — = — I 1 —
J) ^"•""^ r a 0 c
p cos^-4 4- q cob\B + r cos^O = a •\- h + c.
(a) Let AD (fig. 59) be the line bisecting the angle BAG,
i>, D' the points in which it meets BG and the circumscribing
circle respectively.
Draw DG, DH, perpendicular respectively to AB, AC]
then DG = DH = p m\^A,
and D G.AB + DKA G=2 (area of triangle)
= AB.AC sin A',
.'. p sin^-4 {h + c) = be sin J.
= be 2 sin|^^ cos^^ ;
2 cos4^ 1 1
p be
.,. ., , 2cosi^ 1 1
Similarly, ^— = - + - ,
•^ ' q c a
2cosiC 1 1
i. = I •
coshA cos^B cosA(7 111
p) q r a
b-^c
(/3) Again, join BD', CD': these lines will be equal to one
another, since they subtend equal angles at the circumference.
But BD" = c' + p' - 2pc cos^^,
GD'^ = V' + p" - 2pb cos\A ;
.*. c" — 2p'e cos^^ = ¥ — 2p'b cos^^ ;
.•. 2p' cos^-4 = J + c.
Similarly, 2q cos ^5 = c + «,
2r' cos^(7 = rt + i;
.'. p cos^^ + q cos ^5 4 )•' cos| C = a -\- b -{■ c.
104 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
1851.
1. If ABC be a triangle right-angled at (7, E the point
in which the inscribed circle touches BC^ and F the point in
wliich the circle drawni to touch AB and the sides CL4, CB
produced meets CA : shew that if EF be joined, the triangle
FEC is half the triangle ABC.
Let r, / be the radii of the inscribed and escribed circles
respectively ; then, since 0 is a right angle,
CE=r and CF=r\ and
triangle FEC = \rr'
[s — a) [s — h) (s — c)]* {s[s — a) (s — Z>)]*
= i
s
adopting the usual notation,
= i(6"-a) {s-h)
2
= ^ (Z> + c — a) [a-^c — b)
= 1 [a^ + F -{a- h)'] -.' c' = a' + b'\
= ^ah
= I the triangle ABC.
2. Shew that sin/3 sin 7 sin [y — /3) + sin 7 sin a sin (a — 7)
+ sin a slnyS sin {^ — a.) + sin (7 — /3) sin (a - 7) sin (/3 — a) = 0.
We have, in general,
sin^ sin 5 slnC
= ^smA{cos{B-C)-cos[B+C)}
= i{sm{B+C-A) + sm{C+A-B) + sm{A + B-C)
-mi{A + B+C)}.
Hence, if ^ = /3, B=y, C=y-B,
sluyS sin 7 sin (7-/3) = i (sin 2[y — /3) + sln2y8 -'feln27}.
Similarly,
sin7 sina sin (a — 7) = 4 {sin 2 (a — 7) + sin27 — sin2a},
sin a sin^ sin (/3 — a) = ^ {sin 2(/3 — a) + sin 2a — sin 2/3}.
1851.] PLANE TRIGONOMETRY. 105
Again, if ^ = 7 - ^, i? = a - 7, C = /S - a,
sin (7 — /3) sin (a — 7) sin (yS — a)
= i{sin2 (iS - 7) + sin2 (7 - a) + 8in2 (a - yS)} ;
therefore, adding these equations, we get
sinyS sin7 sin(7 — yS) + sin7 sina sin (a — 7) + sina siii/9 sin (/3 — a)
+ sin(7 — ;S) sin (a — 7) sin(/3 — a) = 0.
3. The equation sina; = 0 has not any imaginaiy roots.
We have — ^ 2 smx = e""" — e '^.
Now, every imaginary quantity may be expressed under the
form a H — */3. Substituting, then, this quantity for a*, we get
-Ij: — ix --, -o — i, a
£ — £ =£"£/* — £ ' S.P
= cos a (e"'' - £^) + -* sina {e~i^ + z^) ;
therefore, if sina; = 0, we must have
cosa(£'/^-£/') = 0,
sina(£-/^ + £'^) = 0.
These require, either that
cosa = 0 and £~^ + e'' = 0,
which cannot be satisfied by any real vakie of /3 ; or that
sina = 0, and £~^ - e'' = 0,
which can only be satisfied by yS = 0, shewing that a; = a, a real
quantity : whence the equation sin a; = 0 has not any imaginary
roots.
4. If the cosines of the angles A^ B^ C\ of a plane triangle
be in arithmetical progression, shew that s — a, s — h^ ^ ~ ^1 "^^
be in hannonic progression, s being the semi-siun of the sides.
We have
cosvl = 1-2 sin'^^, cos5 =1-2 sin'^ B, cos C = 1 - 2 sin'^ (7;
therefore, if cos^, cos^, cos 6*, are in arithmetical progression,
sin'*^^, sin'' ^5, sin'^'^C, are so;
lUG .SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
^(^s-a){s-c) _{s-a){s-b) {s-c){s-b)^
ac ab CO
2h c a
+
' s — h s —
c s — a
or 2[^--(5-Z>)] ^ s-[s-c) ^ s-{s-a) ^
s — b s — c s — a ''
2s s s
+
s-b
s — c s — a
s — b s
1 1
+
c s — a
whence , 7 , , are in arithmetical progression ;
.-. s — a , s - b , s — c, are in hannonieal progression.
107
SPHERICAL TRIGONOMETRY.
1848.
1. In a right-angled spherical triangle, shew that
sin a tan^^ — sinb tan^J5 = sin (a — b) :
shew also that if ^ be the spherical excess,
. 1 T-, sinAasini& , „ cosia cos^b
cos^c cos^c
(a). We have in general
, , 1 — C08-4
tan*^ = ; — 5 — ,
^ sin^ '
, . . sin(7 . sina
and sin>4 = — — sma = -; — ,
suic sine
since C is a right angle ;
.•. sina tan^^ = sine (1 — cos^),
= 8inc(l — tan6 cote) by Napier's rules,
= sine — tanZ> cose :
similarly sin 5 tan ^5 = sine — tana cose;
.'. sina tan^^ — sinJ tan ^5 = cose (tana — tanZ*),
cose . , ,,
= 7 sm a — 6 .
cosa cos 6
But by Napier's rules, cose = cosa cos^,
.•. sina tan^^ — sin J tan^^ = sin (a — b).
(^). Again,
sin'^^a sin^^i + cos^'^a cos'^^J = :^{(1 — C08a)(l — cosi)
+ (1 + cosa)(l + cosJ)},
= ^(1 +co8a cos J),
= ^(1 +co8c) by Napier's rules,
= cos"''^o;
108 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
fsin^asin^bV /cos^a cos|/a' • 217;. . 2 if /,\
I - ^_ \ _|. / f — — ±_\ = sin^i!/ + cos^ii...(l) :
V cos^c / V cos^c
ami in any spherical triangle
w^ r,\ cos^ia — b) .^
^^ ' cos^(a4-J)
also .1 + 5 = 180° + ^ - C, hence
cotMC-^) = ^i^|cot|C;
^^ ^ cos(^a + ^>) ^ '
therefore, since C is a right angle,
1 + tan^JS* 1 + tan^a tan ^5
1
— tan^^"
1 — tan^a tan^J '
. tanij5; =
tan^^a tan^J,
sin^^
sin^a 'm\\b cos|rt cos^Z*
cos^^
cos^^c ' cos^c '
),
sin^^ =
sin^a sin^J
cos^c '
cos4^ =
cos^a cos^J
therefore, by (1
cos^c
the required formnlse.
2. If three small circles be inscribed in a spherical triangle,
having each of its angles 120°, so that each touches the other
two as well as two sides of the triangle, prove that the radius of
each of the small circles = 30°, and that the centres of the three
circles coincide with the angular points of the polar triangle.
Let ABC (fig. 60) be the triangle, draw the great circle AD
to D the bisection of ^C; let 0 be the centre of one of the
circles, through 0 draw the great circle EOF perpendicular to
BG and intersecting AD in E^ FE will be a quadrant ; ajso join
^0 by a great circle BOH^ 50^ will bisect the angle ABG\
draw OG the great circle perpendicular to AD. OG and OF
Avill each be equal to r the radius of the small circles : let
BG=2a, BF=x, FD = y.
Then, in the right-angled triangle ABD,
coa A BD = tani5Z> cot ^5;
1848.] SPHERICAL TRIGONOMETRY, 109
or, since L ABD = 120°,
1 1 — tan'' a
— h = tana — — ,
^ 2 tana '
1 — tan''a
2 '
.-. tan'^a = 2,
and cos a = —^ .
Again, in the right-angled triangle OBF^
smBF= cot OBF.tan OF,
1
or smcc = — ^ tanr,
and in the right-angled triangle FO G, OF = 90° — r, and
/ OFG = FD = 2j, and
sin06^ = sinO^sinO^^?,
or sin?' = cos?- sin^,
.•. amy = tanr ;
and cosa = cos(x + j/),
= (1 — sin'''a;)*(l — sin^i/)* — sin.r siny,
or = (1 - ^tan'''>-)4(l - tan'r)* - ^ tanV ;
2
.*. cos'' a -I- rj cosa tan^'r + ^tan*r = 1 — ^tan^r + ^ tan*?-,
or i + I tanV = 1 - | tan^ r,
.*. tan'"'?- = ^,
and r = 30°.
Again, let BOH = h, BO = z- then, in the right-angled
triangle BHCy
cos BC = cosBH.cosHC,
or cos 2a = cos J cosa,
.*. 2 cos"''a — 1 = cosa cosi,
^^ - 3 = 51 C086, •.• cosa = -. ,
and cos5 = — wi)
3* ' • -"^^^-34
1
110 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
also in the right-angled triangle OBF^
sin OF = sin OBF.m\ OB,
3*
or \ = — sin z :
2 2 '
.'. sin2; = — r = — cosA,
3* '
.-. J = 90° + 2;,
and 0/f is a quadrant.
Again, joining OC, we have
cosOC = cosO^ C0SJ5C + slnO^ sin 00 cos OB C,
= cosz cos2« + sinz sin 2a cos 60°,
_ _ 2^ 1 1 2.2* 1
~~3*'3'^P~3~2'
= 0,
and 00 is a quadrant, so is also 0-ff; therefore 0 is the pole
of -40: similarly, if 0', 0" be the other centres, 0' is the pole
of CB, and 0" of BA ; therefore 0 0' 0" is the polar triangle
of ABC.
1849.
1 . If P be the perimeter of a spherical triangle, of which the
angles are A, B, 0, and the spherical excess F, prove that
^.^^^^ [sin^E smjA - ^E) sin {B - ^E) sin ( 0- ^E)]^
^ 2sin^^ sin^iJsin^O
By the expression for the sine of a side of a spherical triangle
in temis of the angles,
{sini^sin(^-i^)sin(J5-i^)sin(0-i^)}i=isinasin5sinO,
= ^smh sinO sin^ = |sinc sin^ sinJ5,
1 2
= ^(sina sini sinc)^(sin^ sin5 sinO)^ (1).
Again,
siniPslnfiP-a))i . (sin(^P-Z>) sin (^P-c)]*
I ^ [ sin 5 sine J '
COsi^ =
^± »un^2.
( smo smc
with similar expressions for the cosines and sines of ^B and ^ 0 ;
cos"!^ cos^'iPcos'-'iO sin^'iP
sini--4 siniP sinAO sin a sin?; sine '
I Hoi.] SPHERICAL TRIOONOMETRY, 111
sinasinisinc ~ sin'^^ sin''^i?sin''^C '
_ sin''^ sin'^i? sin'^6'
. 1 r» 1 / • • 7 • \3 (sin^ sinjB sin (7)
.". sin^i^= i(sma sino smc . , . — . , „ . — rr^,
^ *^ ^ sm^A sm^B sm^C^
_ {smEsmjA - ^E) sm{B - ^E) smjC - ^E)]i
~ 2sln^^ sin^Z? sin^C '
bj (1); the required formula.
1851.
1. If ABC be a spherical triangle, right-angled at C, and
cosvl = (cosrt)^, shew that b + c = ^tt or |7r, according as b
and c are both less or both greater than ^tt.
By Napier's rules
cos^ = cosrt sin 5,
but by the conditions of the problem
cos^ = cos'''a,
.•. cos a = sin 5.
Again, by Napier's rules
cose = cos a cos J,
= sini? cos J from above,
sinJ5 . , ,
= . , smo coso,
smo '
= — — smb cos J, since 0 is a right angle ;
.•. sine cose = sin?> cosJ,
or sin2e = sin 26;
and b is not equal to c, as then B would be a right angle, and
A would equal «, which is contrary to the equation cos^ = cos^/;
^euce 2& + 2c = TT or Stt.
Now b and e are both greater or both less than ^tt, since
cos^ or cos'^a = tan J cote ; therefore 2J + 2e = tt or 37r, according
as b and e are both less or both greater than ^tt.
112
THEORY OF EQUATIONS.
1849.
1. Given y = .rs", prove that
- = 1 + 717 + ^ + ••• +^T^'+ •••
X [2 [3 \n
One root of the equation
?/ - a-c^ = 0
is the coefficient of - m the expansion of — log [ 1 ) .
(See Murphy's Theory of Equations^ p. 77, Art. 62, and p. 80,
Ex. 3.)
Now -log 1 = — +-^ + ...+ -+ ...
V y I y ^ y ^ y
Expanding the exponentials, we see that the coefficient of - is
^x' i'x^ n"-'x
2 1.2.3 1.2 .„ n '
which is therefore a root of the given equation.
\«-i
-u- y ^ 2x i^xY inx)
Hence ^ = i + — + ^-—L + ... + v — L_ + ...
X [2 [3 \n
2. If a^j, x^...x^ be the roots of an algebraical equation,
and no two of them be equal, then
1 1 1 1
jn-1
7n being a positive integer less than n.
1849.] THEORY OF EQUATIONS. 113
(a). Here f{x) = {x-x;){x-x^),..{x-xj,
we may therefore assume
1 A, A A
f{x) X - X^ X - x^
X — X
1
-4j, A^y..A^^ being independent of a;;
.-. l=A^{x-x;){x-x^)..,{x-xJ+A^{x-x;){x-x^)...{x-x^){x-'X^)+...
+ A^^ {x — x^)...{x — a;,^_j) identically ^
Hence, putting x = x^^
1 = A^{x^-x^){x^-x;)...{x^-x„),
= Af{^.) (1);
similarly 1 = AJ'{x^) (2),
i = A/'K) W-
Hence
1 ^ 1 1_ _ 1_ _
/ (^) ~ (•» - a^i) />i) (^ - a'J f\^,) '"'^ ix- xj f{^J
identically.
Therefore putting a; = 0, which makes /(a?) =^„,
J_____l 1__ _ 1
\_ 1 1_^ 1
_ 1 / ^ 2CJ
and similarly for the other fractions. Hence
1 _ 1 / ^ ^ \ _1
H-? + ^+-.l+z>,VJl+f + |{ + ...)
1 /, rr cr' \
+ ... +-rrr-s 1 + --f -^ + ... (1).
1
114 SOLl'TIONS OF SENATE-HOUSE PROBLEMS. [1851.
B^t J_ ^ 1 ^ 1
f{x) x" +2hx-' + ... +2)„ ^n /j ^l\ _^ _^ A
V X '" x^
a; V •>c
Hence, if m be a positive integer, less than «, the coefficient
of ^ in 7^— r = 0 ; and the coefficient in the right-hand member
X J-[x)
of (1) is
1 — J — ? i- -I — « — •
^ ffl-l ^ m-i ^ m-i
hence J. . + ^^tt — r 4- • . . + ■^, — ^ = 0.
1851.
T,^ 1 1 — i^ «
1. If x+ = - 3;?,
1 — X X -^
then (x + y-^ - 0)'-' i^) = - 27 (^; - a>) (^ - a>y,
where (o is an imaginary cube root of unity.
1 1 — X
Call the quantities a;, and ^ , ^/j, j/^? ^^d 3/3 5 then
y^+y, + yz = - ¥ (i),
a^ 1
and y,y, + yjj,^y^, = Yzr^-x~ ^^"^^
= 1 1 + a; - 1
1 — a; X
= y, + y,■^y^-^
= -3(i^ + l) (2),
also y,y^y^ = - 1 (3).
Again,
x^ 1 (1 - a;)"
y^y^ + y^yz + 2/3V1 = j-^r^ - ^^fz^ + ^
_ 3x{\-x) _
~ x{l-x) ~ ^*^'
1H51.] THEOllY OF EQUATIONS, 11/
and y'^y^ + y^y^ + y'^y^ + y'^y^ + y~y^ + y; y^^
= -y^ {3 (p + 1) + ^,3/3} + -^7(2),
= 9p (^; + 1) + 3, by (1) and (3) ;
••• y"y, + y"y. + y'y, = 9i? (i> + 1) + e ... by (4) and (5).
And [y^ + mj.^ + a)'V3)' = j/^ + y/ + ^3"^ + 3^?/,y, (?/, + 6)yJ 4
similar tenns + ^y^y^^
{3/1 + (*" + ^) 3/2} + similar terms
= (-3i^r- 9(&)-l) + 3(<y'''-l)
{9p(^ + l) + 6}by(l), (4), and(5),
= - 27/ + 27/ (26)'' + ft)) - 27^? (2 + w)
+ 27&)''', since a> + &)'•* = — 1,
= - 27 (/> - ft)) (p - ft)7''
12
( no )
GEOMETllY OF TWO DIMENSIONS.
1848.
1. With two conjugate diameters of an ellipse as asymptotes
a pair of conjugate hyperbolas is constructed ; prove that if one
hyperbola touch the ellipse the other will do so likewise ; prove
also that the diameters drawn thi'ough the points of contact are
conjugate to each other.
Let the equation to the ellipse, referred to the conjugate
diameters, be
1^ + ?-=' W-
And to the hyperbolas
^1/ = ^" (2),
^I/ = - c' (3).
(a) In order that (2) may touch (1), we must have
^ _^ , ^
a perfect square, in which case we shall have also
X" xy y^
a c 0
a perfect square, and (3) ^v^ll also touch (1).
(/3) . If the above expressions be perfect squares, we see that
4 _ J_
- ah
••• ' =Y'
and - = ± - ;
X a
1848.] GEOMETRY OF TWO DIMENSIONS. 117
and \i x'y be the coordinates of the pohit where (1) meets (2),
,, he' W
^ =2-
Similarly, if x"i/" be the coordinates of the point, where (1)
meets (3),
„„ b ,,,, a
y'=2^ -^ =2-
And if r\ r", be the lengths of the corresponding semi-
diameters, CO the angle between the axes,
r"' = x'^ + 2^'^ — 2x1/' cos w = ^[d' + b^ — 2ab cos to),
r'"' = x"^ + y'"'' — 2x"y" cos g) = \[c^ + i^ + 2ah cos oj),
... r"' ^r"' = ce + h';
therefore ?•', r" are conjugate to each other.
2. Shew that the curve which trisects the arcs of all seg-
ments of a circle described upon a given base is an hyperbola
whose eccentricity = 2.
Let AB (fig. 61) be the base, a its length,
AC = CD = DB = r, CAB = 0,
we then have r {1 + 2 cos 6) = a,
or, referring the curve to A as origin and AB as axis of a;,
x' -\- y' = [a-2xY'^
.-. 3j?' - / - 4aa7 -|- a' = 0,
the equation to an hyperbola, the squares of whose axes are to
one another in the ratio 3:1, and whose eccentricity therefore
= (3 + 1)^ = 2.
3. Let Z) be a point in the axis-minor of an ellipse whose
eccentricity is e, S the focus, 0 the centre of curvature at the
D 9
extremity of the axis-minor ; with centre D and radius = — —
118 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
describe a circle ; shew tliat this circle will touch the ellipse or
fall entirely without it, according as D is nearer to or further
from the centre than the point 0.
Let C be the centre of the ellipse, and let CD = h^ also let
a, J, be the semi-axes of the ellipse, its equation will be
a o
Also DS' = d'^ + li^^ so that the equation to the circle will be
72
a;"' + [y + ]if = a^ + 4 (taking D below C),
1 - e'
or .-r^ 4- y + 2/^ J/ = «' + K' — ^— .
Where this meets the ellipse, we have
y W aV «■' o' e' "'
or ^-^/ - 2% + A'' -^ = 0;
.-.3/ = -^- h.
If this value of y give a real value for a-, the circle will
touch the ellipse, if not, it will fall entirely without it, since its
radius f ct' + -5 ) is greater than a, and therefore, a fortiori^ than
DB\ which is less than h.
In order that the value of x may be real it is necessary that
y be not greater than Z>, therefore
\-e' ,
e'
or
he'
<-^-h,
?
<B0-
BC\
<C0;
1848.] GEOMETRY OF T\V(» DIMENSIONS. 119
therefore the circle will touch the ellipse, or fall entirely without
it, according as h < or > CO^ i.e. as D is nearer to or further
from the centre than 0.
4. PSp is any focal chord of an ellipse, A the extremity of
the axis-major ; AP^ Ap meet the directrix in two points Q^ q :
shew that I QSq is a right angle.
We may prove this property for any point in the ellipse
by a process exactly similar to that of Part I. Conies^ 1848, 3 ;
except that we have the equations
sin PBS = e sin PSR. sin PRN^
sin QR8 = e sin QRS sin PRN,
instead of those there given.
This theorem may also be proved by the method of Reci-
procal Polars. (See Salmon's Conic Sections^ chap. XIV.)
Take the polar reciprocal of the whole system with regard to
the focus S. To the ellipse w^ill correspond a circle, to the point
P, 2J^ two parallel tangents Rt^ rt\ (fig. 62) variable in position.
To A (or any point in the curve) will coiTespond a fixed tangent
tt\ and to the directrix the centre S'. Hence to AP^ Aj) will
correspond the points t^ t' respectively, and to Q^ q the lines
St^ St'. But it is easy to see that the lines Sf, St' are at right
angles to one another ; therefore the line joining the points Qj j,
subtends a right angle at the focus S.
5. In the given right lines AP^ AQ, (fig. 63) are taken
variable points ^j, q, such that Aj) : pP :: Qq : qA • prove that
the locus of the point of intersection of Pq, Qp is an ellipse,
which touches the given right lines in the points P, Q.
Let AP = o^ AQ = 5, Ap = a, Aq = yS; then the conditions
of the problem give
a:a-oi::h-^:/3, or - + f = 1 (1).
an
120 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
Take AF^ AQ, as axes; then the equations to Pq,pQ respec-
tively, are
M = ' (^).
M- (^)-
o a
whence, eliminating a, yS from (1), we get
X y
0 a
x^ xy if ^ X y
a
ab 1/ a h
the equation to the locus of the intersection of Pq^ p Q^ which,
since the square of half the coefficient of xi/ is less than the pro-
duct of the coefficients of x^ and 3/^, is an ellipse.
When £c = 0, we have
J, J ^ + 1 - 0,
•'• 3/ = ^
shewing that the ellipse touches AP in P.
From considerations of symmetry it is evident that it also
touches AQ m Q.
6. A parallelogram is constructed by di'awlng tangents at
the extremities of two conjugate diameters of an ellipse ; prove
that the diagonals of the parallelogram form a second system
of conjugate diameters, and that the relation between the two
systems is reciprocal.
1848.] GEOMETRY OF TWO DIMENSIONS. 121
Let the equation to the ellipse referred to the conjugate
diameters be
<+C = i (.).
b
The equations to the tangents, drawn at the extremities of
these diameters, are
a; = a, x = — a^
y = h, y = -h]
therefore the equations to the diagonals of the parallelogram
thus formed, are ^ .,
M (^)>
I=-f («)■
At the points where (2) meets (1), we have
a b
therefore the equations to the tangents at these points are
M=±^' W'
therefore these tangents are parallel to (3). Hence the diagonals
fonn a system of conjugate diameters.
Again, the equations to the tangents at the extremities of (3)
M = ±^' («)'
and at the intersection of (4) and (5), we have either
cc = 0, or 3/ = 0,
shewing that the diagonals of the parallelogram, formed by the
lines (4) and (5), are the first system of conjugate diameters;
hence the relation between the systems is reciprocal.
7. PSp is any focal chord of a parabola whose vertex is A,
prove geometrically that AP^ A_p will meet the latus-rectum in
two points Q, q, whose distances from the focus are equal to the
ordinates of the points j7 and P respectively.
122 auLUTIUNS OF SENATE-HOUSE PROBLEMS. [1848.
Draw P3/, prn^ (fig. 64) ortllnates to the points P, p re-
spectively. Then since SQ is parallel to Pil/,
.-. SQ:AS:: MP : AM,
.-. SQ:4.AS'':: MP: ^AS.AM,
:: MP: MP']
.-. SQ.MP=^AS\
Now SP=2A8-\- SM,
= 2A8-\- SPcoaPSM',
.-. SP{l-cosP8M) = 2^^,
.-. Pil/(1 - cosPSM) = 2A8s{nP8M:
similarly pm[l -\- cos 2)8 m) = 2 A8 sinjy 8m j
.'. PM.pm = 4:A8'\
= 8Q.PM from above ;
.-. ^wi = 8Q,
similarly PM = 8q.
Or the distances of Q, q, from the focus are equal to the
ordinates of the points ^j, P respectively.
8. From a given point in a conic section, draw geometrically
two chords at right angles to each other which shall be in a
given ratio.
The construction which we shall give depends on the pro-
perty that all chords of a conic section which subtend a right
angle at a given point P of the curve, intersect the normal at P
in a fixed point.
Draw PK (fig. 65) the normal at P, and draw PU^ PV any
two chords at right angles to one another. Join UV, cutting the
normal in K. Then by the property above enunciated, if PQ, PR
are the required chords, QR will pass through K. Again,
PR
t3in PQR = -^^ a given ratio, hence the angle PQR is known ;
on PK we describe a segment of a circle containing an angle
equal to PQR ; let it cut the ellipse in Q. Join PQ, and di'aw
PR at right angles to it, PQ, PR will be the required chords.
1848.] GEOMETRY OF TWO DIMENSIONS. 123
9. DeteiTnine the equation to the conic section wlilch passes
through five points whose coordinates are given ; and thence
shew that the equation to the conic section which passes through
the five points whose coordinates are
1,-1; 2, 1; -2, 3; 3, 2; -1,-3,
is 6iy - llxy - 65a;^ + 36^/ + 174a; - 151 = 0.
Let ajj, y^ ; x^, y^ ; a-,, y^ ; a„ y^ ; a,, y.^, be the coordinates of
the five given points, which we shall call yl,, A^^ A^j A^j A^ re-
spectively. Then the conic passing thi-ough A^y A^^ A^j A^^
circumscribes the quadrilateral, whose sides are A^A^j -^3^4j
A^A^^ A.A^. The equations to these sides are
^-^2 y-y^ =0
^2
-
^3
X
-
^3
^3
-
^4
X
-
X,
^4
-
^5
X
—
^5
y, - y.
y-yz
y, - y.
= 0,
y-y.
= 0,
y,-y.
y -y.
= 0.
^5 -^2 y,- y^
Now the equation to a conic, circumscribing a quadrilateral,
the equations to whose sides are ii^ = 0, u^ = 0, u^ = 0, u^ = 0
respectively, is ^ ,, ^ -^^.^
2 4 3 0/
\ being an indetenninate parameter.
Hence the equation to a conic passing through A^, A^^ A^j A^^
'^ 'X- x^ y - y\ (X - x^ y-Jh}
■^2 - ^3 yt- yJ '
^^4 - a^6 y,- yJ
fx-x^ y - yy
^fx-x^ y- y.^
v^3 - ^4 3/3 - y,^
' \^S - ^2 3/5 - 3/2.
= \
The quantity X is detennined by the condition of the conic
passing through the point A^{x^y^) ; this gives
'^1 - -^4 y, - y,
^4 - ^5 y*- Vi
'^1-
3_
3/1-
^
'^2-
3/2-
■yJ
r^
-•^3
_^
-y.
y^ - yj v^6 - ^2 .'/.-, - .'/■>
124 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
Eliminating X between tliese last two equations we get,
clearing the quantities within the brackets of fractions,
{[x-x^){y-y,)-[y-y,][x-x,)\\{x-x:^{if-y^)-{tf-y,)[x-x^)]
{(^,-«'2)(3/2-^3)-(3^-2/J(^2-^8)}{(-'^-'^4)(3/-y5)-(yi-3^4)(^-a^6)l
{ (-^ --^a) [y-yy (3/ -3/3) (•^3--^4) 1 { (-^ --^J iy-y-y (i/-y.) i^^-^^)] '
the equation to the required conic. The reduction of this to the
symmetrical form would be very tedious, and we shall therefore
leave it in the above shape.
In the numerical example
^, = I7 3/1 = - 1 ; ^2 = 2, 3/, = 1 ; a-, = - 2, ^3 = 3 ;
^4 = 3, 2/4 = 2; x^ = -l, y^ = -3.
Hence
x^ — x^ = — \ 'y a^j — cCg = 3 ; x^ — x^ = — 2 • x^ — x^ = 2j
3/,-3/2 = -2; y,-y, = -4^; y,-y, = -^] 3/^-3/5 = 2,
a;^ — X3 = 4 ; x^ — x^ = — b-, x^ — x^ = A ', x^ — x^ = - S^
?/.2 - 3/3 = - 2 ; 3/3-3/4 = 1; 3/4-3/5 = 5; 3/5 - 3/2 = - 4 ;
therefore the above equation becomes
{_ 2(a;-2) - 4(3/-!)} {5{x-S) - ^y-2)]
{(-2)(-l)-4(-2)H5(-2)-4(-3)}
|l(^ + 2) - (-5)(y-3)} {-i{x+l) - {-S){y + S)] .
{l(3)-(-5)(-4)}{-4(2)-(-3)2}
(8 - 2a; - 4j/)(- 7 + 5x- 4.y) _ {- \Z + x -\- 6ij){b - Ix + 3j/)
•'• 20 ~ 34 '
.-. 17(5a;-4?/-7)(a; + 2j/-4) - 5(a; + 5?/- 13)(4a;-33/- 5) ;
.-. 65a;' + llxy - 61/ - 174a; - 36y + 151 = 0,
or 61/ - llxy - 65a;' + 36^/ + 174a; - 151 = 0,
is the equation to the conic passing through the five given
points.
10. Two chords AB^ AC are drawn from a given point A
in a cm*ve of the second order so as to contain a given angle,
shew that BC will always touch a curve of the second order.
1H48.] GEOMETRY OF TWO DIMENSIONS. 125
Let the equation to the given conic section, refen'ed to A
as origin, be
Ax^ + 2Bxi/ + Cf + 2l)x + 2Ey = 0 (1),
and let ax + ^y = I (2)
be the equation to -BC, a, /3 being variable parameters.
At the points of intersection of (1) and (2), we have
Ax^ + 2Bxy + Cy' + 2 {Dx + Ey){ax + ^y) = 0,
or ( C + 2E^)y' + 2 (5 + ^a + D^] xy^ + (^ + 2Z>a) x' = 0...(3).
This may be considered as a quadratic in "- , and if ^j, ^2 ^® ^^^
roots, ij, t,^ will be the tangents of the inclinations to the axis
of X of AB^ A C respectively. But AB^ A C include a constant
angle, iarC^m suppose; hence we must have
-^ = m.
1 + tA '
or ^ \,^' r^-^ = irc.
(I + M2)
Now from (3), by the the theory of equations
B+Ea^D^ _A + 2l)a
C+ 2EI3 ' '' C + 2^/3 '
.-. 4i{{B+Ea + I)^Y-{A + 2l)a){C+2E^)]
= m' { {A + 2Da) + C + 2E^]\
.'. 4.[E' - AC-\-2{BE-CD) a + 2{BD-AE) yS + E^a'
+ D'0' + 2{B+DE)a^],
= m'{A + C + 2{I)a + E^)]%
which may be written under the foi-m
aa' + 2hafi + c^' + 2(f7a + e/9) + 1 = 0 (4),
a, 5, c, <7, e being certain determinate functions of A^ B^ (7, Z>, J?,
and in.
Now consider the conic section whose equation is
^V + 2B'xy + Cy + 2[D'x + E'y) + 1 = 0 (.5).
126 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
Where (2) meets this, we have
^V + Wxij + Cy + 2{D'x + E'7/){aa: + ^i/) + [ax + ^i/Y = 0;
.-. {A + 2D' a + a') x' + 2 (5' + E'a + D'^ + ayS) xy
+ {C'-{-2E'^ + ^')f = 0.
In order that (2) may touch (5) the roots of this equation,
considered as a quadi'atic in - , must be equal ; we must therefore
have
[B' + E'a + Z>'/S + a^Y = [A' + 2D' a + a') ( C + 2E'^ + /3'-') ;
... B" - A'C + 2 {B'E' - CD') a + 2 [B'D' - A'E) yS
+ {E"'-C') oi' + 2{B'- D'E) a/3 + [D" - A') ^' = 0,
which agrees with (4) if
E"' - C'_ _ B' - D'E' _ D" - A' _
B^ -A'C'~ ' B" -A'G'~ ' B'' - A C
B'E' - CD' _ B'D' - A'E' _
B"'-A'C ~ ' B'-'-A'C ~^'
which five conditions can be satisfied by means of the five
disposable quantities A'^ B', C, D', E . Hence BC always
touches the conic whose equation is (5).
This theorem may also be proved by the method of reciprocal
polars. For taking the polar reciprocal of the whole system
with regard to ^ ; to the conic will coiTespond a parabola, and
to -S, C (two points the line joining which subtends a constant
angle at the origin) will correspond two tangents containing
a constant angle. The reciprocal theorem then is :
If two tangents be dra\\Ti to a parabola, including a constant
angle, the locus of their point of intersection is a curve of the
second order.*
This may be proved as follows :
* Taking the polar reciprocal of this systena mth regard to the focus of the
parabola, the theorem to be proved is the following :
If a chord of a circle subtend a constant angle at a given point of the curve,
it always touches a circle, which is knoAvn to be true.
1848.] GEOMETRY OF TWO DIMENSIONS. 127
Let y = tx + -
be the equation to any tangent to a parabola. This may be
written ,, „
t--y.t + -^0 (1).
XX
This equation, considered as a quadratic in t^ gives the
tangents of the inclinations to the axis of the two tangents
di'awn to a parabola through a point {xy). In order that these
may include a given angle tan~S?i, we must have, if ij, ^^ be the
roots of (1), t — t
-^ = m ;
1 + f,t,
therefore, by the theory of equations.
1 + ^
X
or y^ — 4acc = nf[a + x)'\
the equation to the locus of xy^ which is therefore a curve of the
second order.
11. Pf J) are the extremities of two semi-conjugate diameters
of an ellipse E^ whose semi-axes are «, b ; upon FD describe
an equilateral triangle PDR^ so that the point R may fall with-
out the ellipse ; the locus of R will be an ellipse E^ : assuming
the above result, shew that if E^ be similarly treated, as also all
the successive ellipses, the axes A^^ B^ of the a;'^ ellipse E^ so
described will be comprised in the fonnula
[a + 5)(coti7r)-' ± (a - J)(cot^7r)*".
In the figm-e (66), let CP^ CD represent the equal semi-
conjugate diameters of the ellipse E^ and let D' be the other
extremity of the diameter through D. Join PD^ PD'\ on them
describe the equilateral triangles PDR^ PD'R ; then i?, R' will be
the extremities of the axes of E^. Join CR^ CR'. Then
CR = ^A^, CR = \B^:
128 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
also ^, = t&nPCR = -, and FV'.CV = ^ab;
CV a' ^ '
.-. OF' = i«, and PF' = ij.
And CR' = CV + V'B',
= _ + 2iZ>cos-.
O TT
Similarly Ci? = ^ + 2*a cos - ,
2* 6
l(^, + JJ = (« + &)(2*cos^ + i),
AT ^X 7^ 1 3^+1 2*
Now 2*cos-+-, = -^^ = 3P33
3^+l\| /2 + 3hi /l + cosi7r\|
3^-1/ V2-3V Vl-cos^TT/'
= lcot^)*.
Similarly it may be shewn that
2^cos|-i = (tan^, ,
.-. i(5, + A) = (« + ^)(cot^),
i(5,-^J = (a-J)(tan^y,
a formula connecting the axes of any two successive ellipses ;
.-. ^^ = (« + 5)(cot^) 4- («_J)(^tan^) ,
A = (« + ^) (cot ^) - (a - &) (tan ^) .
1849.] GEOMETRY OF TWO DIMENSIONS. 129
Now assume
B^ = {a + b) (cot ^)*' - -^ (« - b) (tan ^)'",
A=(« + ^)(cot^) +-'(«-^)(tan^) ;
tlieu, by the formula already given,
^^^ = -—T-" (cot - ) + ^.r-^ (tan-) ,
2 V 127 2 V 127
= (« + J)(cot-j +_-(a-J)(tan^j ;
•Ml A A + B / 7r\*- A - B f ir\^
similarly J,,, = ^-^ (^cot - j + ---^^ (^tan -j ,
= (« + J)(cot^) +_-(«- J) (tan ^j .
If then the assumed fonn hold for E^ it is proved to hold for
^^j. But it has been shewn to hold for E^^ therefore it holds
universally.
1849.
A is the origm (fig. 67), 5 a point in the axis of?/, BQ a line
parallel to the axis of ar ; in AQ (produced if necessary) P is
taken such that its ordmate is equal to ^^: shew that the locus
of P is a parabola.
Let AB = a, AP = r, BAP = ^ir - 6^ then the ordinate
of P= rsin^, also BQ = «cot^;
.'. r sin^ = acot^,
or r^ siv^ 6 = ar cos 6 ;
therefore, putting r sin 0 = y^ r cos^ = a-,
y' = c^,
shewing that the locus of P is a parabola,
a® b^
2. K from points of the curve — ^ + -a = (a* — b^j tangents
be drawn to the ellipse -7 + 75 = 1, the chords of contact will
'^ a b
be normal to the ellipse.
K
130 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
If
at the
1, 7] be the
point (.r^),
current coordinates of the normal
its equation will be
to
the
ellipse
X
a'
v-y
" y '
01
r -
a^ 1,
T2 2 ' •
a" — h^ X 1/ — (^ y
If ar^, y^ be the coordinates of the pole of this line with
respect to the given ellipse, we have
_ g" 1
- ^' 1
y^ - y^ -a^ y-
Eliminating jc, y between these two equations and the
equation to the ellipse, we get
? + p = («■'-*')■''
the equation to the locus of the pole of the normals to the
ellipse.
Hence if from points of this curve, tangents be di'awn to the
given ellipse, the chords of contact will be normal to the ellipse.
3. An oblique cone stands on a circular base; prove that
one of the axes of the section made by a plane passing through
the centre of the base and pei-pendicular to the axis, is a mean
proportional between the other axis and i)seca, where D is the
diameter of the base, and a the angle between the axis and a
normal to the base.
Let MPM' (fig. 68) be the section through 0 the centre of
the base and perpendicular to CO the axis of the cone ; let the
section intersect the circular section RPR' in the line NP.
Let a and h be the axes of the section, then
F _ NP' _ RN.NR'
a' ~ MN.NM' ~ MN.NM' '
1849.] GEOMETRY OF TWO DIMENSIONS. 131
Let lCAO = ^, CBO = y; also the z MO A is given equal
to a. Hence
BN _ sin(/3 + a) NR' _ sin(/3'-a) ,
MN~ sin/3 ' :?^~ sm 13' '
Z)"'' _ sin (/3 + a) sin(y8' — a) /-^
'' ^~ sin/3 • siu^Q^ ^ ^'
and a = JL¥' = .1/0+ OJ/'=^l-.-^S^ + _.J^l ..(2).
2(sm(/3 + a) sm(/3'-a)]
Again, since AO = OB,
sm AC 0 sinBCO
sinCAO ~ smOBC
cos (8 + a) cos f/S' — a)
or ^; — pj-^ = — \ — 7^7 — ^ = p suppose ;
snip smp -^ ^^
^ ;:> + sina ,^, ^ - sina
.•. cGt/3=^^— — -, cota =~ .
cos a cos a
Substituting in equations (1) and (2), we have
12
-jj = (cosa + cotyS sin a) (cos a — cot/3' sina),
(1 +2^ sina)(l —]) sin'a).
cos a
_ 1 — ^^sin"''a^
cos'^ a '
, D I cosa cosa
and a = — -. y t , .
2 VI +p sin a 1 — ^?sina/
I) cosa
1 — p sm a
.-. = jr, i>seca,
or h'' = aD sec a,
and 5 is a mean proportional between a and Z^seca.
4. Let P,, P^, Pj,, (),, (2.,? ^3 be six points lying in a conic
section; let the areas of the triangles P^Cl^Q^, ^iQaQ^i ^iQiQ^i
be denoted by ^,, P,, O,, and the areas of the triangles formed
K2
132 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
by putting /!,, P^ successively in the place of P, be denoted by
^'11 ^27 ^2) -^3) A) ^3 I'espectlvcly ; then will
A,\B,C, B,CJ ■ A^\B,C\ BfiJ ' A,\B,G, B,C„
Let Mj, Mj, Wg denote the distances of any point from the lines
^•2^35 ^3^15 QiQ;^ respectively, then ?/, = 0, tt,^ = 0, u^ = 0 will
be the equations to these lines themselves. And the equation to
any conic section passing through Q^^ Q^^ Q^^ may be written
under the form
^ + ^^ + -^ = 0 (1),
W, M, «3
Aj, X,^? \ being constants whose values will be determined by
the condition of the conic passing through P^, P^, Pg.
Wg', ^3' be the values of u^^ w.^, u^ respectively at Pj,
Let u
u.
u
Then A = 4^.^3-<, A = ^^3^x-<, ^, = i^x^.-<,
with similar expressions for ^.^, P^) ^2 5 -^35 ^35 ^3 •
J <'? < ^j
1 5<"5<" ^-
1 \ 1
\BA
-
i^.cj +
1
A
U<^2
1
8
1' f
1
1 \
" ^2^3-a^,
'Q,qA<\^.
^3
^2
"<'i
1 / 1
% ^
1
7J^.,"
Ct
1
>3"
«2«*2/i
3 ^
And since
P.P^Pg all lie
in (1),
we
have
= 0,
A/, A.„ A„ _
-V + -T, + -7, =0,
%, M, M3
A.. A-„ A<- ^
-7^ + -^, + -^ = 0 ;
i*. W., t*„
1849.] GEOMETRY OF TWO DIMENSIONS. 133
whence eliminating "^^W by cross-multiplication, we get
1/1 1 \ 1 / 1 1
+
1/1 1 ,
+ — , -7^ - -^r-. =0;
whence dividing by ^Q.,Q^.Q^Q,.Q,Q^,
1/1 1\ 1/1 1\ 1/1 1,
5. The equations of three straight lines are
u (= X sin 6 — 1/ cos ^ + c) = 0, «, = 0, m^^ = 0 ;
prove that the equations of the four circles, to each of which
these lines are tangents, are
. . 0-$^ . . 0- 0^ 0 0
w* sm -^— — i + ?/j* sm — - — ^ + u^ sm -i— — = 0,
a a i
w* sm -^-- — - + w * cos ? + ?/ * cos — = 0,
2 2 2
* ^. - ^1 I ' Q-Q. * ^, - ^
w* cos -^^—r — i + w * sm — - — ^ + ^C cos -^ = 0,
2 2 2
* ^2 - ^1 . i ^ - ^. * • ^, - ^ .
M* cos -^-— — J + u^ COS — - — ^ H- ?<2* sm -^ = 0.
^ ^ ^
The equation to any conic section, touched by these three
lines, may be wi-Itten,
\u^ + \u^ + \u} = 0 (1).
K we reduce this equation to the form of an equation of the
second degree, the coefficient of the terms of two dimensions
will not contain c, c^, c.^; hence, to find the condition of this repre-
sentmg a circle, suppose c, c^, c^ to be Indefinitely diminished, the
ratios c : c^ : c^ remaining unaltered.
The three lines u = 0, u^ = 0, ?/,^ = 0, will then all pass
through the origin, and the circle touching them will degenerate
into the origin ; Its equation will therefore be
x^ + if = 0,
or 1/ = ± — * .r.
134 s()h:ti()ns of senate-house problems. [1849.
Equation (1) will therefore become, dividlug out by a;,
\ (sin d + -i cos (9)4 + \ (sin 6^ + -* cos ^J*
+ \ (sin e^ + -4 cos ^Ji = 0 (2)*
Hence, by Demoivre's theorem, equating real and imaginaiy
parts separately to zero,
ff 0 ff
X sin - + \ sin -^ + \ sin ^ = 0,
2 2 2 ( ^^y
e e 0
\ cos - + X, COS -~ + \, cos -^ = 0.
2 2 -^ 2
Eliminating X, X^, X,^ by cross-multiplication from (1), (3),
Ave get
1 • ^., ~ ^, i. • G ~ ^., i • ^1 ~ ^ /x
w* sm -^'— : — - + w,^ sm — - — - + uj sm -^— — = 0.
2 ' 2 ^ 2
Now if in this equation we write tt + ^ for ^, tt + ^^ for 6^,
and rr + 6,^ for 0^, successively, by which substitution equation
(2) is not altered, we get the equations to the remainmg circles :
these are
X • 0.^ — ^, 1 0 — 0„ 1 0,-0 ^
tt* sm -~- — ^ + M * cos — - — ^ + uj cos ^ — 0,
2 2 2
u^ cos
0-0^ ^ . 6' - ^, , ^ - ^
2
-^ 4- M 4 sin — - — - + u} cos -^ — = 0,
"^ 2 2
, 0 — 0 1 0 — 0 i ' 0 — 0
w* cos -^— — - + Wj* cos ^ + Mjj* sm -^-— — = 0,
ia i^ ^
the required equations to the circles.
1850.
1. If at a given point two circles intersect and their centres
lie upon two lines at right angles to each other through the
* This method of investigating the condition that equation (1) may repre-
sent a circle, is due to Mr. Leslie Ellis. It may be shewn in precisely the
same manner, that if ip (m, u^, u^) = 0 be any equation of the second degree, the
condition that this may represent a circle is </> (^"^e, £-i9i, t~i6j^—o.
1850.] GEOMETRY OF TWO DIMENSIONS. 135
point; prove that, whatever be the magnitude of the circles,
their common tangents will always meet in one of two straight
lines which pass through the given point.
Take the given point as origin, and the lines on which the
centres lie as axes. Let a, )8, be the radii of the circles. Then
the intersection of the common tangents must always lie on the
line joining the centres of the circles, whose equation is
? + ^=l.
From considerations of symmetry it is easy to see, that if this
intersection always lie on one of two fixed lines passing through
the origin, the equations to these lines must be x + y = 0,
X — y = 0. Hence, if such be the case, the equation to one of
the common tangents must be
X y X + y
- 4- I + ^ -1=0,
a ^ 7
where 7 is a constant to be determined.
In order that this line may touch the circle whose radius is a,
it is necessary and sufficient that its distance from the centre of
the circle, whose coordinates are a, 0, be a. We must there-
fore have
1 i\' /I 1 Y" '
a 7/ Vy9 7/
.}__(]_ ly (I 1
' ■ 7' ~ Va 7/ V/3 7
shewing that 7 is a symmetrical function of a and yS, and there-
fore that if this straight line touch one of the circles, it must also
touch the other. Hence the intersection of the common tangents
always lies on the line a; + y = 0, if a and yS have the same
sign, i.e. if both centres lie on the positive, or both on the nega-
tive side of the origin. If one centre lie on the positive, the
other on the negative side, similar reasoning will shew that the
intersection of the common tangents lies on the line x — ?/ = 0.
136 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
2. A number n of equal confocal parabolas are ranged all
round the focus at equal angular intervals ; shew that the product
of the distances of all the points of intersection from the focus
is — 5— , I beinff the latus-rectum.
Taking the common focus as pole, the equations to the para-
bolas will be
' = -b ("'
4 sm -
2
-—77B~^ <^''
V2 ^ n)
* ™ (2 + it)
4 sm - H TT
V2 n
If ^j, 0^ be the two values of 6 at the intersections of (1) and
{m + 1) it is manifest that since these intersections lie at the
extremities of the same chord passing through the pole,
Also we easily see that u, = ,
.'. C/ = TT .
^ n
Hence, if r^, r^ be the corresponding values of ?•,
/ I
' . , mTT ' ■■^ , ., mTT '
4 sm -— 4 cos — -
271 2n
• •. rr = — .
^ ^ . „ mTT
4 sm —
n
1850.] GEOMETRY OF TWO DIMENSIONS. 137
Similar expressions holding for the intersection of (1) with
each of the other parabolas, we have
product of all the distances of intersections of (1) with the other
parabolas
^ ' Sin — sni — sm tt
w n n
-r, , . TT . 27r . n — \ n
rJut sm - Sin — ... sin tt = -^, ;
n n n 2
^2(«-l)
therefore the above product = - — 5— .
n
To get the product of all the distances of intersections, we
n
have merely to raise this quantity to the power - : (not n ; since
each intersection would then be counted twice over) ;
therefore product of all the distances of intersections = — s- •
3. The locus of the points from which a circle is projected
into a circle, upon a plane inclined at a finite angle to that
of the given circle, is an equilateral hyperbola.
Let 0 (fig. 69) be the centre of the given circle, AB that
diameter of it in which it is cut by a plane through 0, pei-pen-
dicular to the line of intersection of the plane of the circle, and
the plane of projection. Let CD be the corresponding diameter
of the circle in which it is projected. Join CA^ DB^ and
produce them to meet in E^ E will be the point from which
the given circle is projected.
Draw EG parallel to CD and equal to AB^ terminated by
EC^ ED : let EG, AB intersect in P, then must AP = GP^
BP = FP. Through 0 draw two lines OX, 0 F, parallel to
those respectively bisecting the angles APE, APG, and take
them as axes. Let «, h be the coordinates of P; — «, — 5 of ^ ;
X, y those of E. Then x will be the abscissa of P, and it is
hence casv to sec that n -\- 2.r, — h will be the coordinates of G.
138 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
Hence, f , 77 being cuiTent coordinates, the equation of ED will be
f — a 7} — h
a + 2x — a — h — b^
^ — a 7} — b
or -= — —
(1)
X b
V /
And tlie equation to EP is ^ = x
(2)
At E^ the intersection of these two lines, we have f} = V-, and
from equations (1), (2),
x~b'
.'. xy = aJ,
shewing that the locus of E is an equilateral hyperbola, of which
OX^ 0 Y are the asymptotes.
4. Prove that y = nx -\ ad iniinitum is the
^ X + X + ... ''
equation of a hyperbola. Find the position and magnitude of
the axes, and write down the equation of the conjugate hyper-
bola under the same form.
Since y = 7ix +
y — nx =
X -{- X -\-
1 1
X + X +...
1
X -Y y — nx^
- 1
y — [n—l] x^
••• {^J-nx){y-{7i-l)x] = 1 (1),
the equation to an hyperbola, whose asymptotes are represented
by the equations , , ^
"^ ^ y = nx^ y = (7? — 1) x.
The axes bisect the angles between the asymptotes, therefore
l.SaO.] GEUMETliY UF TWO DIMENSIONS. 139
their equations are
y — nx _ y — [n—\)x ,
{\ + ny " {1 + {n-\y\^ ^^^'
y — nx y — i^n—\)x
(Tm?? " " {i + (/i-ir}i ^^^
respectively. Hence the positions of the axes are known.
To find the magnitudes of the axes, we have, combining (1) (2),
y — nx = ,
{l + (n-iyf
y - [n-l)x = ^ 5^ ^ ;
(i+«'r
. {i+(^-iri^-(i+^-o^
• • »^ — ' •
_n{l + (n-l)']^- (n-l)(l + ny
{l-flw-irni+n")'
= 2 [(1 + 71^)4 [1 + {n- lY]i - [n' - n + 1)],
which gives the square of the magnitude of one of the semi-axes.
Similarly the square of the other semi-axis may be shewn to be
= 2 [(1 + riy {1 + (n - 1)^}4 + {n' -n + 1)].
The equation to the conjugate hyperbola is
{y-nx)[y-{n-\)x] =-1,
which may be written
1
11 — nx — ,
^ X ■\- y — nx^
_ 1 1
y
X — X —
1 1
.r — X — ..
is the equation to the conjugate hyperbola.
140 SOLUTIONS OP SENATE-HOUSE PROBLEMS. [1850.
5. From a point 0 (fig. 70) are drawn two lines to touch
a parabola in the points P and Q\ another line touches the
parabola in R and intersects OP, OQ in 8^ T; if V be the
intersection of the lines joining PT^ QS crosswise, 0, i?, V are
in the same straight line.
Let OP = a, OQ = h, then the equation of the parabola
referred to these lines as axes, is
!)' + (!/- (')•
Let 08= oi, 0T= /3, then the equation to 8T will be
M- c^)-
To find the condition that (2) may touch (1), we proceed as
follows :
Square each member of (1) and multiply it crosswise by (2),
we thus get
X y _ f/^"\- (y
a "^ ^ " |W "^ V^
This may be considered as a quadratic in \ ] •, and in order
that (2) may touch (1), it is necessary that its roots be equal ;
hence its first member must be a perfect square.
The equations to PT, Q8 respectively, are
- + ^ = 1
a b
where these meet, we have
the equation to a line passing through the origin, and through
the intersection of PP, Q8^ that is to OV.
185U.] GEOMETRY OF TWO DIMENSIONS. 141
Again, to find the equation to OR^ we have since the first
member of (3) has been made a perfect square,
therefore squaring, ^ (" " -) = 2^ (jg - -^j '
'1 1\ /I r
^^Kl--^)+K^-^)=''
the equation to OR^ which agrees with that ah-eady formed for
0 V] hence 0, i?, V are in the same straight line.
6. A series of circles pass through a given point 0, have
their centres in a line OA^ and meet another line AB. From
ilf, iV, the points in which one of the circles meets the lines
OA^ AB^ are drawn parallels to AB^ OA, intersecting in P.
Shew that the locus of P is a hyperbola, which becomes a
parabola w^hen the two lines are at right angles.
Take 0 as origin, OA as axis of x, let the equation to any
one of the circles be
x^ +f = 2rx (1),
and that to AB xcosol + ?/siua = a (2) ;
hence the coordinates of M are 2/-, 0, and the equation to the
line through J/ parallel to AB^ is
xcosa + ?/sina = 2rcosa (3).
Now the circle in general cuts AB in two points, either of
which may be denoted by N. The equation to the line through
either of these points parallel to OA (the axis of ic), will be ob-
tained by putting the ordinate of that point = 0, and therefore
the equation to the pair of parallels will be obtained by elimi-
nating X between (1) and (2). This gives
(« -y sina)'"* + y' cos' a = 2r cos a (a — 3/ sin a) ;
.-. y'^ + 2 (rcosa — a) ^sina + d^ — 2arcosa = 0...(4).
To find the equation to the locus of P, the intersection of (3)
with either of these lines, we must eliminate r between (3) and
142 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
(4), whence we get
y^ + (iccosa + ^sina — 2a) ?/sma + d^ — a (a?cosa + y slna) = 0,
or ^■'(1 +8in'*a) + xi/ cosa sin a — a [x cos a, + 3^ sin a) + d^ = 0.
This is the equation to the locus of P which is evidently in
general an hyperbola. If however the lines are at nght angles
a = 0, and the above equation becomes
y'^ — ax + d^ = 0,
representing a parabola.
7. If from the focus of a parabola, lines be drawn to meet
the tangents at a constant angle, the locus of the points of inter-
section will be that tangent to the parabola whose inclination
to the axis is equal to the given angle. Prove this in any
manner, and shew that if m be eliminated between y = mx -\ — ,
and V = [x — a), the result contains a factor which
answers to the locus. Also explain briefly the origin and
signification of the other factors.
(a). Let 4a be the latus-rectum of the parabola, a the iji-
clination to the axis of any one of the series of tangents, yS the
constant angle at which the lines through the focus meet the
tangents. Then taking the focus as pole, and the axis as
initial line, the equation to the tangent is
r = acoseca cosec(^+ a).
That to the line through the focus is
6/ = 7r-(a + ;8).
Eliminating a, we get as the locus of the intersection of these
"^^^ r = acosecyS cosec(^ + yS),
representing the tangent whose inclination to the axis = /3.
/3. From the equation
m + t .
y = [x — a) (1),
we get m [x — a + ty) = — t{x — a) + y ',
1850.] GEOMETRY OF TWO DIMENSIONS. 143
therefore combiuing this with
y = mjc + ^^ (2),
y — tix — a] fy + X — a
y = ^ !^ 1 X + — -. -, a,
ty + X — a y — t[x — a)
y— tix—a) , , ill— tix—a) ty + x — a
- '^ ^ {x-a)+a]^ ^ ^ + ^
ty + x-a^ [ty + x — a y — t[x—a))^
{{x-a)^ + f}{l+f) .
ty + x — a) [y — t[x — a)} '
X — a)'' ■+ y' ty — fx — a
^f+[x-ar^^ {[x-aY + f]{l+f) .
ty + x — a i^ty + x — a)[y — t[x — a)]^
= 0,
X — a + ty ' y — t{x — a)
This is satisfied by y = tx + - ^
representing the locus found above.
It is also satisfied by
[x-aY + y' = %
which requires that cc = «, ?/ = 0, representing the focus.
This would be obtained by making m = (—1)* in equation (2).
Its signification therefore is, that if tangents be drawn to the
imaginary branch of the parabola, got by making x negative,
and lines be drawn through the focus of the real branch cutting
these tangents at a constant angle, the point of intersection of
these lines will only be real when the tangent to the imaginaiy
branch of the parabola passes through the focus of the real
branch.
8. Within the evolute of an ellipse is inscribed a similar
ellipse ; within its evolute another similar ellipse, and so on
ad infinitum ; shew that the sura of all the areas
TT
id'+V')
4 ah
Let mrt, mh be the semi-axis of the first inscribed ellipse,
then m will be a homogeneous function of a and h of no dimen-
sions, and therefore the same function of ma and mh ; hence
144 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [18oU.
wt^/, ni^b will be those of the second, and the sum of all the areas
TTuh
1 - m' '
we have therefore only to find the value of m.
Now the equation to the evolute of the ellipse is
{axf+{hyf = {d^-hi (1).
In order that this may touch the ellipse
\ina) \mh)
they must have a common tangent at a common point.
Now the equation to the tangent to (1) at [xy] is
In order that this may touch (2) at {xy)^ we must have
'a^i 1 X
(S^^.-(3.. = K-0-'.
2 7
27,a J
X ina
' • 2 2 2 z3j 7
ma a — 0
y^ _ mW'
therefore, by (2), m = ^r-^ 5
therefore if S equals the sum of all the areas,
Trab
S =
a'- b'\'
a' + ¥
4 ah
1850.] GEOMETRY OF TWO DIMENSIONS. 145
9. Find the points ^,, A^^ A^...A^^^_^, A^^^ in a parabola, such
that the tangents at these points are parallel to the focal dis-
tances SA^^^^ SA^, SA^y..SA^_^^ ^^„t-\-) respectively.
TiQl A^A^KV [^g. 71) represent the parabola, A''/S'X its axis,
A^T^^ ... A^T^^ the tangents at ^„...^,,, respectively, and let
lA^SX = a^, then lA^T^X = ^a^. Hence, by the conditions of
the problem, we must have
W=^m (1),
K = «. (2),
K = a,-, (''),
^m-l = am-2 (w-1).
The last equation will be
TT + ia,,, = «„,_, (m),
since this will satisfy the condition of A^^^T^^ being parallel to
SA^^^_^^ and it is manifestly inconsistent with the preceding
equations to have ^a^^^ = a^^^_^.
Hence, multiplying generally equation (?') by 2'""^, and add-
ing all equations thus formed, the quantities a^, a^, ... «,,_ , dis-
appear, and we get
^ + i««, = 2"'-'a,„ ;
- 27r
•■• «m - 2'" _ 1 •
whence ., = ,-J^ ,
2'-+«_.
and generally, a,. = ^„. _ ^ ,
and the positions of the points A^^ A^...A^ are determined.
10. From the focus of an ellipse lines are di-awn to any four
points in the curve, and the reciprocal of each line Is multiplied
by the sines of half the angles between any two of the remaining
lines ; prove that the sum of the first and third of these products
taken in order is equal to the sum of the second and fourth.
L
146 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
Let L = the latus-rcctum, e the eccentricity of the ellipse ;
then Its equation, refeiTcd to the focus as origin and the axis-
major as prime radius, will be
1 _ 1 — e cos^
and let ^,, 6^^ 6^^ 6^ be the angles between the axis and the
successive lines, r^, ;-^, rg, r^ the lengths of the lines.
Then — sm -2-— — -* sm ^— — ? sm -^ ^
y-j 2 2 2
2 ^0-0, 0-6. (^-f>
= ^ (1 — e cos ^j) sin -^-— — * sm -^-— — ? sm
Z ^ >^ 2 2 2
= ^ (1 - e cos ^J (sin [6^ - e,) + sin (^, - ^J + sin (^3 - ^,)}.
O- ., , 1 . ^, - ^. • ^.> - ^4 • ^4-^1
bimilarly — sm ^-- — sm -^— - — sm -^-— — -
•' r^ 2 2 2
= ^ (1 - e cos^3) {sin (^, - e^) + sin (^, - ^J + sin(^^ - ^J];
therefore if ^8' denote the sum of these two quantities, and If the
symmetrical fmiction sin {6^ — 6^ + sin {6^ — 6^ + sin (^3 — 6^
+ sin (^.^ - 6^) -}- sin [d^ - ^J + sin [6^ - 6^ be denoted by <^,
and if we put c^ for cos^^, s^ for sin ^j, &c.,
+ ^3 k^l - ^1^.2 + ^4^2 - ^^4 + ^'^1^4 - ^4^1)}]
= ^ [0 - &A[^-h)+ ^-f^i^-^l) + C3^4(«l- «2)+^4^lK-*3)}]-
Similarly, if S' be the sum of the second and fourth products,
^S" will be obtained from S by writing 6,^ for 6^^ 0^ for ^.^, 6^ for ^3,
^j for 6^ ; hence (/), which Is a spmuetrical function, will remain
unchanged, and
whence It appears that S = /S",
or the sum of the first and third of these products is equal to the
sum of the second and fourth.
1851.] GEOMETRY OF TWO 1)1 .M KXSloNS. 147
11. If lines be drawn tlirough any two of the points
A^ i?, Cy.. and other lines through any two of the points
f/, i, Cy.. all in one plane, prove that the intersections of AB
with «7>, of AG with «c..., will all lie in one straight line, pro-
vided that the lines through the intersections of any two of
the first series of Hues and the corresponding intersections of
the second series all pass through the same point.
Conceive the points -4, i?, C, ... to lie in one plane, and
((,b,Cy.. in another; then, since the lines joining the intersec-
tions of any two of the first series of lines and the corresponding-
intersections of the second scries all pass through one point,
Aa^ Bbj Cc... all pass through one point 0.
Now consider any quadrilateral, as ABah^ whose angular
points are any two points of the first series, and the correspond-
ing two of the second series. Since the lines Aa^ Bb intersect,
they are in the same plane, therefore also AB^ ab^ are in the
same plane, and must therefore intersect,* and their intersection
must manifestly lie in the line of intersection of the planes
ABC..., abc... Similarly the intersection of any other pair of
lines, as A C, ac, lies in that line.
Hence, if we suppose the planes ABC..., abc..., to be indefinitely
nearly coincident, the proposition enunciated follows at once.
1851.
1. Having given a focus and two tangents of a conic section,
shew by means of reciprocal polars, or otherwise, that the
chord of contact always passes through a fixed point.
Let a circle be described passing through two fixed points,
A, B, and let F be the intersection of the tangents at A, B.
The locus of P will be a fixed straight line, pci-pendicular to and
bisecting AB.
Now take the polar reciprocal of this system with respect
to any fixed point S. The reciprocal of the circle will be a
conic section whose focus is S, and which has two fixed tangents
(the reciprocals of A, B). Hence the reciprocal of P, which is
* If these lines happen to be parallel we may still consider them as inter-
secting in a point infinitely distant.
L2
148 SOUTTIOXS OK SENATE-HOUSE PROBLEMS. [1851.
tli(^ chord of contact of these tangents, will always pass through
a fixed point, the reciprocal of the locus of P.
2. Shew that there will be two pairs of equilateral hyper-
bolae which pass through two given points -4, B^ and touch two
given straight lines, and that the chords of contact of each pair
meet in AB^ and are equally inclined to AB.^
Take the middle point of AB as origin, AB as axis of a*,
let h^ — h^ be the abscissse of A^ B, respectively, and let the
equations to the two given tangents be
5 + f_l = 0, ^+1-1=0,
and that to their chord of contact - + ^—1=0,
a p '
where a, /9, are indeterminate parameters.
Then the equation to a conic section touching the two given
lines may be written mider the form
M->)(M-')=^(M-'
\ being an indeterminate parameter.
Two equations for the determination of the three arbitrary
quantities X, a, yS, are given by the conditions of its passing
through u4, B. We thus get
^o(^')=Ks-y «-
a J \a J \a.
The third equation is given by the condition of the curve
being an equilateral hyperbola. In order that this may be the
case, it is necessary that the sum of the coefficients of x^ and
y-* = 0. This gives
— - ^. + 777 - 3i = 0 3 .
aa a. bo p
* A shorter solution of this problem, due to Mr. Gaskin, will be found in
the Appendix.
1851.] GEUMETUY OF TWO DlMENS-lONS. 149
Combining (1) and (2), we get
^h — aV (h — a) (h — a)
Kh + aj (A -f a) {h + a) '
This is a quadratic for the determination of a. Let a,, a„ be
its roots.
Subtracting (1) from (2) and eliminating \ by (3), we get
1 1\
n i\
2
/I 1 ■
,? + w)
\a a'J
a
\aa' bh\
Hence, for each vahie of a there will be two of yS, equal and
of opposite signs. Let them be denoted by ^8^, — ^,, ^^^ — /S^
respectively, then we get two pairs of equilateral hyperbolae,
whose chords of contact respectively are
- + !- = 1, --i = 1 5.
It is easy to see that the lines represented by (4) intersect
in AB (the axis of x) and are equally inclined to AB. The
same will be the case with the pair of lines denoted by (5).
Hence the chords of contact of each pair meet in AB^ and
are equally inclined to AB.
3. If from a point of an ellipse a line be drawn to the ex-
tremity of each axis, and a parallel to the same axis be drawn
through the point in which such Hue meets the other axis, the
locus of the intersection of these parallels is an equilateral hy-
perbola.
Trace the coiTesponding positions of the point on the hy-
perbola, and the point on the elhpse.
Let a, J, be the semi-axes of the ellipse, take the axes of
the cui've as coordinate axes, and let a cos a, h sin a, (a being a
variable parameter) be the coordinates of the point tlu'ough
which the lines are drawn. The equation to the line drawn
through this point to the extremity of the axis-major, is
X — a cosa y — ^ sin a
a cosa —a h sina '
150 S<lMTIttN8 OF SKNATK-IIUUSE PROBLEMS. [1851.
X — a cosa y — h sin a
or ^-j +'' =0;
a sill ^ a h cos^a
X a 11 . OL a
.'. - cos- + Y sill- = cos - .
a 2 h 2 2
Where this meets the axis-minor,
ij = hcot^a. (1),
which is therefore the equation to the line through this point,
parallel to the axis-major.
Again, the equation to the line drawn through (« cosa, b sin a)
to the extremity of the axis-minor, is
y — h siiia _x — a cosa
Z»sina — h a cosa '
y — h cos (I TT — a) x — a sin ( 2 "^ — a) _
h cos(-2 7r — a) — h asin(^7r — a)
whence, by an investigation similar to the above, it is seen that
the equation to the line drawn parallel to the axis-minor through
the point where this meets the axis-major, is
X = acot(j7r — |-a)
=«3^i (^i-
Eliminating a between (1) and (2), we get
y + b
y-h'
or xy = ay + hx + ab^
or {x — a) [y — b) = 2aJ,
as the equation to the locus, which is evidently an equilateral
hyperbola, whose asymptotes are the tangents to the ellipse at
A and B.
Let ABA'B (fig. 72) be the ellipse, P the point (a cosa,
b sin a) ; then from the figure it appears that p is the point
on the hyperbola corresponding to P: one branch of the hyper-
bola is described, while P moves from A io B'^ the other branch
while 2> moves round through BA'B'A.
1851.] GEOMETRY OF TWO DIMENSIONS. 151
4. Dctenniue the values of a', m\ n\ such that the relation
{{x-af + /}* = m (x^ + y'y- + n
may be equivalent to the relation
[{x - a)" + y'Y- = m {x' + y'^Y + n.
The transfonnation fails (1) in the case where the curve
represented by the given equation is a conic section, (2) has a
double point.
(a). The equation
{{x-a'Y + f]^ = in {x' + f)^ + n\
when transformed to polar coordinates, and rationalized, becomes
r^ — 2a.'r cos^ + a.'^ = (?«';• + ?«y,
or (1 -m")r' - 2a'r cos^ - 2m'n'r + a' - ?i" = 0.
The equation
{(a; -a)' + /}* = m {x'+iff- + n,
similarly transformed, becomes
(1 - 7n') r^ - 2a.r cos 6 — 2mnr + ai' — n" = 0.
In order that these equations may be identical, the coefficients
of r^, r cos ^, r, must bear the same ratio to one another as the
constant temis,
1 — ni'' OL m'n a^ — n^
1 - m^
a
mil a'"* -
-n'
lation
a
^
a" - n"'
Q ■>
a
a^ - 71^
1-^
we get
a'
a
=
-?
, »"
'-^'
1 -
1 - in'
I -
n'
^
1 - m' '
^'
152 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Also wc have
n
a
mil 7)1
a
'e
—
^
, or —
— ^ i
a
vin VI
n
71 '
a
•
m'
^
a '
and
n'
=
m,
a
a
iiice
a
1 - nf '
1 a' - n'
•
a!
^^
a 1 — m^
711 a"'* — n^
n
^
a 1 — nf '"
(1),
(2),
(3).
The required values of a', m', w', are determined from (1), (2),
and (3).
(y8). The given equation, when rationalized, will in general
be of the fourth degree in x and y. In order therefore that
it may represent a conic section, it is necessary that the terms
of a degree higher than the second should disappear of them-
selves from the rationalized equation, which requires that
7)1 = ± 1. If this condition be satisfied, the given equation
DGCOmGS
{{x-ay + f]i± {x'+7/'y = 7i,
shewing that one of the foci of the conic section is the origin,
that the coordinates of tlie other are a, 0, and that the axis-
major = 71.
The transformed equation must therefore, since it represents
a conic section, take the form
{{x-a'Y + 7/}i±{x'+f)i = 7l\
which gives a', 0 as the coordinates of the second focus, 7i as the
axis-major. Hence we must have
a' = a, w' = ;«,
and the transformation fails.
1851.] GEOMETRY OF TWO DIMENSIONS. 153
The transfonnation also fails if n = a, for we then get m' = 1 ,
a = 0, n = 0, and the transformed equation becomes an identity.
In this case, the given equation, transformed into polar coor-
dinates, becomes
(1 - vf) r - 2 (a cos ^ + ma) = 0 ;
whence we see that if
cos^ = — m^
we have r = 0 ; but the equation cos 0 = — m is satisfied in
general by two values of ^, whose sum = 27r ; hence the origin
is a double point.
If therefore the curve have a double point, the transfonnation
fails agam.
5. If 0 be the centre of a reflecting circle, Q a radiant point,
and the line from Q to 0 produced to meet the circle be con-
sidered as the axis, then, if a be the radius, u the distance QO^
0 the inclination to tlie axis of the radius through the point of
incidence of any ray, and <f> the inclination to the axis of the
reflected ray,
pcos(f) = acosd + wcos2^, psmcj) = asm0 + usm20j
where p = {a' + u^ + 2au cos 0Y is the length of the incident ray.
Let P (fig. 73) be the point of incidence of any ray, M the
point in which the reflected ray cuts the axis. Let
and draw PN perpendicular to the axis. Then
QN= /3 cos (<^ - 21/r) = QO + ON = u + «cos^,
PN = p sin (0 — 2\|r) = a sin 0.
Again, yjr = (ji — 0,
.'. p coa{'20 — (j>) = u + acos0 (1),
p8in(20-<^) = asin^ (2):
(1) cos 2^ + (2) sin 2^ gives
p cos<^ = a cos 0 + u cos 2^,
154 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
(1) sill 2^ — (2) cos2^ gives
psincf) = aamO + usm20j
the required equations.
6. Using the notation of the last question, and assuming the
tiTith of the theorem stated therein, shew that if fi-om the point
of incidence of each ray there be drawm, in a direction opposite
to that of the reflected ray, a line equal in length to the incident
ray, the locus of the extremities of these lines is a curve cutting
the lines at right angles, and the equation of which, referred to
the radiant point as origin and the axis QO SiS axis of x^ is
x^ + y^ — 2ux = 2a{x^ -i- y^)^.
Shew that the origin is a double point, and trace the curve :
shew also that the equation may be expressed in the form
{{x — af + /]* = m [x'-\-y^)^ + n.
' (a). Produce MP to i?, making PR = QP^ then we have
to find the locus of R. Let x^ y be its coordinates, then we
readily see that
x= QN -\- pcos<f)
= u + acosd + acosd + ncos20
= u{l + cos2d) + 2a cos ^,
and y = PN + psin^
= asin6' + asin^ + ?/ sin2^
= M sin 2^ + 2a sin 6 :
hence - = tan^, and RQ is parallel to PO,
and [x-
-uY
+ f
= u^ + 4«'' + 4:au cos 6,
.-. x' +
■f-
-2ux
X
■— A /y 1 A /ytt
x' + f -
2ux
+ li'
x' . ■> . X
^ +
w.^
x''
2 , -2 — *<^* "T ^:''" / 2 , 2^
X +y G-^ +,y .
0.'^+/'
1851.] GEOMETRY OF TWO DIMENSIONS. 155
.'. x^ + }f -2ux = 2rt(x'+y')*,
the required equation to the curve.
(y8). This equation, transformed to polar coordinates, becomes
r — 2u 0,09,6 = 2a,
.-. r = 211CO&0 + 2a (1).
Hence the radius vector of the curve exceeds by 2a the
radius vector of a circle which passes through the pole, one
of whose diameters is prime radius, and whose radius = u. If
therefore we draw such a circle, and produce 0-4*, the radius
vector of any point A in it to i?, making AB = 2a, the locus
of R will be the required cm've.
The cm-ve will pass through the origin when cos^ = ,
which condition, if a < u, is satisfied by two values of 0, one less,
the other greater than tt. Hence if a < w, i.e. if Q be outside
the circle, two branches of the curve pass through the origin,
which is therefore a double point.
When ^ = 0, /■ = 2 {u + «), and when 6 = tt, r = — 2 [u — a) ;
hence the curve will have the form represented in fig. (74), where
Qq = 2{zi + a), Qq =2{u-a).-f
Again, l QRM= L 0PM (since QR is parallel to OP) yfr = (f>-0,
sm<f> cos^ — cos^ sin^
tSinQRM =
COS0 cos^ + sin<^ sin^
u sin 0
a + Mcosi
by the result of question 5.
* Tlie line OA must ahvaj's be produced in the positive direction of the
radius vector, therefore when 0 > .Vtt, OA must be produced backwards.
t This is the form of the figure when Q is outside the circle : if it be witliin
it, the curve docs not pass through the origin, and the loop Qq' docs not
appear. The origin will then be a conjugate point.
15G SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
And if L be the angle between the radius vector and tangent,
1 flu u sin 0
cott = — T7i = —
L dO a + u cos 6 '
= tan()i?J/.
Hence the curve cuts the lines PR at right angles.
(7). The equation
[{x - a)' + y'Y- = m {x' + y^ + n
is equivalent to
(1 — m^)r^ — 2mnr — 2ar cos^ + a^ — w^ = 0,
and this coincides with (1), if
-2 = «? -. -2 = ^1 a. - n =0]
1 _ ,n^ - "' 1 _ ni'
a u — a
m = - . 71 = a =
,/<- — >*— ,
u u
therefore (1) is equivalent to
,2 2^ 2 N J
w — a\ ,J^ u , ., .,,, It — a
'■^\*
^ — :— +3/ =-i^"+rf +
)
M / j a u
which is in the required fonm.
7. Given the centres of three circles, each of them touching
the other two externally, determine the radii.
How many systems of circles are there when the centres
are given, but the circles touch externally or internally at
pleasm'c ?
Let a, &, c, be the distances between the given centres ; then
'•2 + ''3 = ^)
^3 + '\ = h
r^ + r.^ = c,
b + c — a
similarly r^ =
2 '
c + a — b
2 '
a + b — c
which determine the radii.
1851.] GEOMETRY OF TWO DIMENSIONS. 157
If the circles touch internally or externally at pleasure, there
will be four systems. For the circle described with any one
point as centre may include the other two, which must touch
each other externally, thus giving three systems. Or they may
all touch externally, giving four systems in all.
8. The locus of the point from which two given circles
subtend equal angles is a circle.
Let -4, A' (fig. 75) be the centres of the two given circles,
P a point from which the circles subtend equal angles. Draw
the tangents P T, Ft to the circle whose centre is A ; PT\ Ft' to
that whose centre is A'. Join P/1, PA\ AT^ At, A'T\ A't'.
Then LTFt = AT Ft'.
And AF, A'F respectively bisect the angles PPf, TFt',
.-. z AFT = L AFT:
also the right angle ATF = \k\e, right angle A'T'F', therefore
the triangles TAF, TA'F are similar, therefore
AF'.A'F:: AT: AT (1).
Divide AA in 0, so that AO \ A 0 :: AT : AT take 0 as
origin, A OA as axis of x. Let AO = a, A 0 = a', and let
X, y be the coordinates of P. Then by (1)
... d'%x^a)^^f]=d^[{x-d)'^f],
or (a — a')[x^ -\-y'^) — 'iaax = 0 ;
shewing that the locus of P is a circle passing through 0.
9. The lines joining the corresponding points of two similar
and similarly situated figures in the same plane intersect in a
point.
All sections of a conical surface of any degree by parallel
planes are similar and similarly situated figures, and every gene-
rating line passes through coiTCsponding points. Hence, con-
versely, the lines joining coiTCsponding points of two similar and
similarly situated figures in parallel planes, pass through one
point (the vertex of the conical sm'face of which they are
158 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
sections). Let the planes l)e now made indefinitely nearly
coincident, and the proposition enunciated follows at once.
10. Given any three of the four lines Ox^ Oy^ 0])^ Oq^
(fij^. 76), the foiu'th may be determined, such that if «, h be the
points in which a line tln-ouf2,h a point Q in Oq intersect Ox^ Oy^
and n\ h' the points in which another line through the same
point Q intersects Ox^ Oy^ the point of intersection of the lines
ah' and ah lies on the line Oj).
Let II = 0, V = 0, be the equations to any two lines passing
through the point 0, and let u — \v^ u = Xv, u = X^r, w = \v,
be the equations to Ox, Oy, Oq^j Oq, respectively. Also let
It) = 0 be the equation to Qa, and n — \ii — fiic = 0 that to Qa.
Then the equation
a{u — \v) — [u — \gV — [xw] = 0,
where a is a disposable parameter, represents a line passing
through a. In order that this may pass through h, the above
equation must be identical with
l3{u-\v)-w = 0,
y8 being also a disposable quantity. In order that these equa-
tions may be identical, we must have
a — 1 a\,. — \a
and (Xy — \j){u — \v) — {\y — \.) (« — \v — fiw) = 0,
or {\ - \){u - Xv) + (\ -X^)fiw = 0,
is the equation to ah. Similarly
(\ - \) (^ - K^) + {\ - \) /^^^' = 0,
is that to ah'. Where these intersect, we have
(X^ - \,) [u - Xv) + (X - \) (u - Xv) = 0,
or {X^ + Xy-2\)u + X,{X^ + Xy)v = 0 (1).
In order that this may lie in the line 0/j, whose equation is
ti - X^v = 0 (2)
1851.] GEOMETRY OF TWO DIMENSIONS. 159
(1) and (2) must be identical ; hence
\ i\ + \- 2\) + \ {\ + \) = 0,
or {\ + X,]{\ + \) - 2\X, = 0,
an equation from which, when any three of the quantities
X^, X^, \ , X^, are given, the fourth may be determined, so
that when any three of the lines Ojc, Oy^ Op^ Oq are given, the
fourth may be determined so as to satisfy the required conditions.
11. The radii vectores from the focus of a conic section to
two points of the curve make equal angles with the line drawn
from the focus to the point of intersection of the tangents at the
two points.
Let a, (B be the angles which the radii vectores respectively
make with the axis-major, then the polar equations to the
tangents, referred to the focus as pole and the axis-major as
prime radius, will be
1 2
- = - {^cos^ + cos(^ — a)},
1 2
- = y {e cos^ + cos [6 - y8)|.
r i
Where these meet, we must have
which will be the equation to the line through the focus and the
intersection of the tangents, which evidently bisects the angle
between the radii vectores. Hence the proposition is tnie.
For a demonstration of this theorem by the method of re-
ciprocal polars, see Salmon's Conic Sections^ chap. xiv.
12. If two triangles be circumscribed about a conic section,
their angular points lie in another conic section.
Let u = 0, V = 0, t« = 0, be the equations to the sides of one
triangle, and let the sides of the other triangle respectively
opposite to these be represented by
?/ + &,t' + Cj?c=0.,.(l), a^« + t' + r,?r=0...(2), a^u + \v+ir=0...{?,).
160 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Then, since these two triangles are circumscribed about a
conic section, it will follow that if (ly^'J denote the line joining
the intersection of v = 0 and (3) with that of i« = 0 and (2),
with similar notation for the other corresponding lines, {v^io^),
{^o^v^), (?//',) all pass through one point.
Now the equation to (^'g^t'J is
V 10
w + - + — = 0
that to 2v,u„ IS — -\- V + — = 0
U V
to t/.-y,, — I h w = 0
(A).
The elimination of w, r, lo between these equations would
give the necessary condition that the three lines denoted by
them should pass through one point, or that the two triangles
should be cii'cmnscribed about a conic section.
Now the equation to any conic circumscribing the triangle
(123) can be put into the form
a [a^u -\-v-\- c^w) {a^u + h^v + w) + ^ [a^u -\- b^v + w) [u -{■ h^v + c^w)
+ 'y[u-\- l\v + c^w]{a^u -{- V -\- c,^ic) = 0 (4).
Here the coefficient of li^ is proportional to
« + -+-,
ofv^to ^ + /S + f ,
n -2. . ^ /3
of 10 to h — + 7.
c c
^1 2
Hence, if we give to a, yS, 7 respectively the values which
w, V, w have at the intersection of the lines (A), each of these
coefficients will vanish, and equation (4) will be reduced to one
involving vw^ wu, uv only ; It will therefore also represent
a conic circmnscribing the triangle whose sides are u = 0,
V = Oj 10 = Oj and consequently, if two triangles be circum-
1851.] GEOMETRY OF TWO DIMENSIONS. 161
scribed about a conic section, their angular points lie in another
conic section.*
From the above proof it is not difficult to see that the
converse (which is also the reciprocal theorem) is true.
13. If the angles ^, cj)' are connected by the equation
cos)u, = cos^ cos<^' — sin^ sin^'(l —c^ sin"''/i)-,
and sin0, sin^' arc the abscissae of points on an ellipse, the
semiaxes of which arc 1, (I— c'^)^, then the tangents at these
points meet in a point, the locus of which is an ellipse confocal
with the given ellipse.
Let f , r) be the coordinates of the intersection of the tangents,
then the equation to its polar is
1 — c*
Let sin^, (1 — o'')* cos^, be the coordinates of the points
where this line meets the given ellipse, then
^ . _ 77 cos ^
the roots of this equation in 6 are <^, ^'.
It may be written in the forai
(|sin^-ir = ^^(l-sin^^);
1 -'
.•. sm<p.sm(p = .]
^ ^ 1 - c"
Again, it may be written in the form
( t; cos ^ ) ^, . , , ^.
1-r .
cos ^. cos <^' =
r + ^
• Another solution of this and of several cognate problems, will be found
in a paper by Mr. Heam, in the Cambridge and Dublin Mathematical Journal,
vol. IV. p. 2G5, entitled " Singular Application of Geometry of Three Dimen-
sions to a Plane Problem."
M
1G2 SOLUTIONS OF SKXATE-HOUSE PKOBLEMS. [1851.
therefore, Ity the given equation,
cos/, (f 4- ^) = 1 - f - (] - ^^,) (1 - c^ sin»S
or ^' (1 -t- cos /a) + —^ {cos // - (1 - c' sin»^j = 1 - (1 - c' sin">)S
the equation to the locus of (^, t;,) which is therefore an ellipse,
the squares of wliosc semiaxes are respectively
l-(l-c''sin»^ l-(l-c''sinV)^ ,
1 + cos /x ' ^ ' cos /i — (1 — c"* sin'V)- '
1 — cos/A — (1 — cos/i) (1 — c*^ sin'^yu-)* ^
or ; g ,
sm /i
1 — c'' sin'^/i — cos/u. — (1 — cos /a) (1 — c"* sin"''/u,)* ^
siu'^/i '
therefore, if c' be the distance from its centre to its focus,
,a _ 1 — cos yu, — (1 — cos /a) (1 — (? sin^/i)*
C — ;: — o
sm fx
1 — cos/A - c'"^ sin'''/u, — (1 — cosyu,) (1 — c^sin^/i)*
sin'"' /A
••• C' = «5
whence the ellipses are confocal.
14. If cc, y^ z, r«, are linear functions of the coordinates of
any point, such that no three of the lines represented by the
equations « = 0, 2/ = 0, z = 0, ?i; = 0, meet in a point, the
equation w + [yzf + [zx]^ + [xy)^ = 0 is that of a cui've of the
fourth order having three double tangents, x = 0, y = 0, z = Oj
and three double points, y = z = w, z — x = w^x=y = w.
Shew also that the six points of contact of the double tangents
lie in a conic section.
Where the line a? = 0 meets the cun'^e,
w + [yzY- + {zx)^ + {xy)^ = 0 (1);
we have also w + [yz]^ = 0, or w'' = yz , (2).
1851. J GEOMETRY OF TWO DIMENSIONS. 163
It hence appears tliat the line x = 0 meets (1) only in the
points where it meets the conic (2), that is, in two points only.
But (1) when rationalized takes the form
(«<?'■' — yz — zx — xyY = ^^xyz [x-Vy-k-z- Iw) (3),
mider which form we see that the curve represented by it is
of the fourth order; therefore the line a; = 0 must meet it in
four real, coincident, or imaginary points. Therefore, cither
each of the pomts in which a- = 0 meets (1) must be a double
point, or a; = 0 must be a double tangent to (1) ; for from con-
siderations of symmetry it is clear that both points must be of
the same nature. Also we see that ^ = 0, 2 = 0 stand in ex-
actly the same relation to (1) as a? = 0 does.
Now consider the conic whose equation is
xc^ — yz — zx — xy = 0 (4) .
From (I), (2) we see that all the points in which x = 0,
y = 0, s = 0, meet (1) lie in this conic. Hence these points
cannot be double points, for if they were, a cm've of the second
order would intersect a cm've of the fourth order in twelve
points (coinciding two by two), which is impossible. Therefore
X = 0^ y = 0, 2 = 0, are double tangents to (1).
Again, if in (3) we put x = u', it becomes
{x [x — y — z] — yz\- = 4.xyz {y + z — x),
which can be reduced to
X [x-y-z] +yz = 0 (5),
shewing that the line x = w meets the given curve only in
the points in which it meets the conic (5), that is in two points
only. Hence, either x = w is a double tangent, or it must meet
the cun'c in two double points.
Now at the points where it meets the curve, we have, as
may be seen from (5), z = x = ?r, x = y = lo respectively.
Hence, where the line x = tv meets the cm'vc, it also meets
either the line y = lo or z = iv^ and from considerations of
symmetry, if x = to touch the ciu-ve, y = iv and z = w muM
do so likewise at the same points.
m2
1()4 SOLUTIONS OF SKNATi:- HOUSE PROBLEMS. [1851.
Now we have already shewn that if x = w do not touch the
curve (I), it must meet in two double points, and we have just
now proved that if it does touch the curve, y = lo and z = w
touch it, each at one of the points where x = w docs. Therefore
two tangents can be drawoi at these points in different direction?,
therefore they must be double points. Hence, in either case,
the points in question are double points, viz. z = x = to and
X = y = w\ and similarly, it may be shewn that y = z = w \9
a double point.
15. (a). Describe a circle when two tangents are given,
and a point from which a pair of tangents drawn to the circle
shall include a given angle.
[^). By means of the properties of reciprocal polars, or
otherwise, construct a conic section, when the focus, two points,
and the angle between the asymptotes are given.
(a). Let AB^ AC (fig. 77) be the two tangents, P the point
from which a pair of tangents are to include a given angle a. Let
lBA C = 13. Bisect the angle BA C by the straight hue ^D,
join PA^ and divide it in E^ so that PE : EA : : sin ^/3 : sin ^a.
Also produce PA to F^ so that PF: ^i^: : sin^/S : sin ^a.
Bisect EF in O^ and with G as centre and GE as radius,
describe a circle, cutting AD in H. H shall be the centre of
the required circle.
For since PE : EA : : PF : FA : : sin^/3 : sin ^a, the locus
of a point, the ratio of whose distances from P and A
= sln^/3 : sin^a is the circle of which EF is a diameter.
Therefore, joming PH, PH : AHw sin^/3 : sinia.
Draw PK a tangent to the circle whose centre is H and
which touches AB^ A C, then if r be the radius of that circle,
-^ = sin EPK, -^ = sin 1/3 ;
.-. sin HPK : sin ^/3::AH:PII
: : sin^a : sin ^/3j
HPK
a.
2'^1
ISoL] QEOMETltY OF TWO DIMENSIONS. 165
and a pair of tangents to the circle drawn from P include the
required angle a.
{13). If we take the polar reciprocal of the above system
with respect to P, to the circle will correspond an hyperbola
whose focus is P; to the two given tangents AB^ AG coitc-
spond two given points, and to the tangent through P, including
a given angle, correspond two points on the cm'vc at an infinite
distance subtending a given angle at the focus or (since the
points arc infinitely distant) at the centre. This angle therefore
is the angle between the asymptotes. Kence (/8) is the polar
reciprocal of (a); and therefore the requii'ed conic section may
be constructed by means of the circle there determined.
16. Let P be any point in a conic section whose focus is 8
and eccentricity e ; in >SPtake SQ = \L (the semi-latus-rectum);
draw QR^ ST perpendicular to ;SP, meeting the tangent at F
in R and T respectively; also draw SY perpendicular to the
tangent meeting it in 1^; and let PZ7, QZ drawn parallel to the
transverse axis meet 8Y in U and Z respectively : it is required
to prove one of the following properties :
(1) P is a point in the latus-rectum.
(2) QR passes through the point U.
(3) PU=e.PS. (4) SR = e.ST. (5) SY.SZ={^L)\
(1). Let the inclination of SP to the axis-major be a, then
the polar equation to the tangent at P will be
1 2
- = -^{(cosi9 + cos(^-a)};
and that to QR, r = \L sec (^ — a).
At P, the point of intersection of these lines, we must have
^ = |7r;
therefore R is a point in the latus-rectum.
(2). The equation to /ST Is
/, sin a
tan 0 = ,
e + cosa
16G SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
or ill rectangular coordinates,
jc ama = y {e + cos a).
Now the ordinate of P is
i^L sin a
1 + e cos a '
therefore the equation to PU is
^L sin a
y =
1 + e cosa
At U the point of intersection of these, we have
_ ^L (e 4- cosa)
1 + e cosa '
.'. X cosa + y sin a = ^L.
Now this is the rectangular equation to QB ; hence QR passes
through the point U.
, V m, 1 • n r^- \L cosa
(3). The abscissa of P is .
^ ' 1 + e cosa
That of CThas been shewn to be
^L (e-fcosa)
1 + e cosa '
and P, ?7, have the same ordinate ; hence
PU==e.PS.
(4). Since i2 is a point in the latus-rectum, SR is perpen-
dicular to PZ7, and ST is perpendicular to 6T, Pii to SY^
whence it readily follows that the triangles STR, SPU are
similar ;
.-. PU'.PS:: SR: ST',
but PU=e.PS,
.: SR = €.ST.
(5). The polar equation to QZ is
r sin ^ = ^L sin a,
1851.] GEOMETRY OF TWO DIMENSIONS. 167
and that to ^F is
zj sin a
e + cosa
. ^ sina
.*. sma =
;H-2e cosa + e')*'
therefore at Z^ the intersection of these lines,
r = 6'Z = ^X (1 + 2c cosa + e')*.
Also SY=^\L ^
{(e + cosa)'^ + sin* a}*
1 ^
(1+26 cosa 4- e')*'
.-. 8Y.8Z=[\L)\
==^L
17. If -4j, A^...A^/y rtj, «2---'^i5 ^^ ^^^^ angular points of two
polygons of n sides each, which circumscribe a given circle, and
Pj, P^...F^ the points of intersection of their first, second... h"'
sides respectively ; shew that
Shew also, by means of projective properties or otherwise, that
the same equation is true when any conic section is substituted
for a circle.
From 0, the centre of the circle, draw perpendiculars Oi?,,
OB^...OB,^, 0\, 0\...0h^,, on the sides A^A^, A^A^...A^A^,
a,a^^ ^,«2*'-^n-i^«? respectively. Through 0 draw any line OX,
and let generally B,OX = a„ h^OX = /3, (fig. 78). Then
A^OX = i (a,_, + aj, .-. AOB, = A,OB,_^ = ^ (a, - a,J.
Similarly, a, OB, = M^r - A-J,
and P,C>Z=i(a, + ^,);
.-. P,0j5, = P^OX - B^OX =h{0r- aJ-
Also PA, = a (tan ^. OP. + tan P, OA" ) ,
a being the radius of the circle,
= a{tan^(a^ - a,_J + tan^(|S, - a,.)|
_^^ sin^(/3, - a,_J
cos-^ (a^ - a^_,) cos ^ (/3^ - aj *
168 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Also F.l = a (tan yl,.,, OB,. - tan P, OB)
= a {tan^ (a,.,, - a,) - tan^ (/3, - a,,)}
= a sin^(a,^, -^,)
The expressions for P^a,. and P^a^+j, will be got from these
by simply interchanging a and y9 ;
. Pr, -n sin^(a. -/3,_J
• • ^^''^ - "" eosi (^, - /3,_J cosi (a, - ^J '
Hence
cos^(/?,.^j - ySJ cosi(a, - ^,.
P.^...P.a.
PA..^,.Pa
^ sin I (^, - a,_,) sin ^ (a, - ^,._, ) cos | («,^^ - g,.) cos ^ (^,.^, - ^ J ,
sin^(^,,^j-a,.)sin^(a,.^j-/3,.)cos-^(a,-a,_,)cos^(^,.-^,_J
From the form of this expression it is easy to see, that if we
give r every value from 1 to n inclusive and multiply the re-
sulting fractions together (observing that instead of a,.^j, /3,.^,, we
write otj/SJ, every factor will appear both in the numerator and
denominator. Hence
P,^,.P,a,.P,^,.P,a,...PA-i^A _ 1
PA,P.a,^PAs'^.%'"KA-PA '
or P^A^.P^a^.P^A,^.P^a^. . .P^.^^A
jf f T> i 1 .-. . OP,..OA,..smP,.OA,
It tor P,.A, we substitute ^^^ -, and make
similar substitution for each of the other lines, each member of
this equation will, since OB^ = a, be divisible by
OA,. Oa,. 0P\ . . 0A„. Oa. op:
and there will remain merely a relation between the sines of
angles subtended at 0. The property just proved must there-
1851.] GEOMETRY OF TWO DIMENSIONS. 169
fore be tnie for any figure into which the circle can be projected,
that is for any conic section. (See the article on the Method of
Projections, in Salmon's Conic Sections^ Chap. XIV.)
18. Prove one of the two follo%viiig properties.
(1). When one of the foci of a conic section and two tan-
gents are given, the locus of the other focus is a straight line.
(2). When the centre of the conic section and two tangents
are given, the locus of the focus is an equilateral hyperbola.
The proof of these theorems depends on the property, that
the product of the pei'pendiculars from the foci on the tangent
at any point of a conic section is constant and equal to the
square of the semiaxis minor.
(1). Let J.P, AQ (fig. 79) be the two given tangents, S the
given focus, H that whose locus is to be foimd. Draw /S'F, HZ
perpendicular to AP\ SY\ HZ' io AQ'^ then
SY.HZ= ST. HZ',
.-. HZ'.HZ:: ST : SY,
a constant ratio ; therefore the locus of His a straight line.
(2). Take the centre as origin, and let the equations to the
given tangents be
cc cosa + y s'moL — a = 0 (1), a^cosa' + i/ sina' — «' = 0 (2).
Let I, 77, be the coordinates of one focus, then — ^, — v^ will
be those of the other. Now the length of the perpendicular
from I, 7], to (1) is
^ cosa + T) sina — a;
similarly, that from — ^, — 77, is
— I cosa — 77 sina — a.
Hence we get
(I cosa + 77 sina)'"' - n' = ^\
/3 being the semiaxis minor.
Similarly it may be shewn that
(^cosa' -f- 77 sina'V^ — «'^ = ^'-^
170 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
.•. (^ cosa + 7} sina)"'' — a" = (| cosa' + rj siiia')"' — «'^
or (I cosa + r) sina)'^ — (^ cosa' + r/ sina')"'' = «■* — « '^
the equation to the locus of |, t;, which is therefore a rectangular
hyperbola, the equations to whose asymptotes are
I (cos a + cos a') + t; (sin a + sin a') = 0,
f (cosa — cosa') + ■j? (sina — sina') = 0.
171
DIFFERENTIAL CALCULUS.
1848.
Find the equation to that involute of a cycloid which passes
through the cusp, and shew that in the immediate neighl)om'hood
of the cusp it becomes the curve 2a{4:yY = {3x)*y a being the
radius of the generating circle.
Let X, y be the coordinates of any point in the cycloid
referred to the cusp as origm, and base as axis of ic, s its dis-
tance measured along the arc from the cusp ; ^, rj those of the
corresponding point in the involute. The equation to the
cycloid will be
X = a(^ — sin^),
y = a[\ — cos^),
and we have, since the tangent at {pcy) passes through (f?;) at
a distance s from [xy]^
^ dx
dy
dx
Now 32 = rt(l — cos^) = 2asiii''^^,
do
-^= asin^ = 2a8ini^ cos^;
da
dd ^ '
and s = 4a(l - cos|^) ;
.-. f = a(^-8in6') - 4a(l-co8^^) sin^^,
= a(^ + sin^-4sm^^),
7) = a{l — cos^) — 4a (I — cos^^) cos^^,
= «(3 + cos^ - Icos^^),
the equations to the involute.
172 SOLUTIONS OF SENATE-HOUSK PROBLEMS. [1849.
In the iiiinicdiatc neighbourhood of the origin where 6 is
small, these become
and 77 = a j 4 — -
6' .
+ 24-^
f-.f;
6"
= %.4-
Hence, eliminating 6^
a'--
=(f)'
or
the required curve.
■ 2«(477)^
= m%
1849.
1. K P be a point in a cycloid, and 0 the con-esponding
position of the centre of the generating circle, shew that PO
touches another cycloid of half the dimensions.
Let a(^ — sin^), a(l— cos^) be the coordinates of P, as in
the last problem ; then a6 and a will be those of 0.
The equation to PO is
X — aO y — a
sm V cos V
or X cos Q -^ y ^VQ.Q — a{6 cos Q + sin 6).
Differentiating this equation with respect to Q as variable
parameter,
— a; sin ^ + ?/ cos ^ = a (2 cos Q — ^ sin &) ;
.*. a; = a(^ — sin^ cos^),
= ia(26'-sin2^) (1),
and ?/ — r; ( 1 + cos'* &)
= \a\\ +cos^^) (2).
1849.] DIFFERENTIAL CALCULUS. 173
Equations (1) and (2) shew that the line OP always touches
a cycloid whose cusp coincides with the cusp of tlie original
cycloid, and generated by a circle of half the size of its gene-
rating circle.
2. Find the locus of the ultimate intersections of the lines
defined by the equation
a;cos3^ + ?/sin3^ = a(cos2^)* (1),
where 6 is the variable parameter.
Differentiating (1) with respect to ^,
a;sin3^ - ?/cos3^ = asin2^ (cos^)* (2).
Squaring (1) and (2), and adding,
x' J^ f = d'cos2d (3).
Again, (2) -i- (1) gives
ccsin3^ — ?/cos3^ , ^^
— z — ^ . ,^ = tan 2^,
a;cos3c7 + ysm'do
or — ^— — — — = tan 2^;
1 + tan3^^
X
.-. ^= tan^,
X
1 1 /«\ -1 -1 2 1 — tan*'^
and by (3) x' -\- i/ = a' -
= a
+ tan'
x'^-f
d' + f
or {x^ + yr-^^\^'-f)^
the equation to Bcmouilli's Lemniscate.
3. If e be the eccentricity of a conic section, r the distance
of any point from the focus, p the radius of curvature at that
point, and ds an element of the arc of the curve, then
, Jr' , fd'r
' =d?-^PW'
174 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
Let the equation to the coiiie section referred to its focus as
origin and axis-major as axis of a?, be
x' + f = [ex + cY ;
.*. r = rx + c,
dr
, dr _d'C _ e
ds ds J ('hf\
dx \ \ dxj
d'^r d dr I
ds' dx ds ' ds '
dx
e^ it
dx'' dx 1
-||)T'I'
dy^
e dx
"^-(IJ
drV , fdS'Y , ^ + \dx)
dlX
ds) ^P Uv ~'' 7±V'
"^ [dxJ
= e\
4. If u be a function of the independent variables x, ;/, z^
given by the equations
^*=/(^, 0 (1),
s = F{Ix + w??/ + nz + ht)
= 0 [mz — ny) + % [nx^ ~Iz) + yjr [ly — mx)^
and if V + m' + n' = F ; shew that
, du du du , du
aa; dy dz dt
where -j- is obtained from (1) by considering s constant.
1850.] DIFFERENTIAL CALCULUS. 176
We have
also =z n')(^ — wi/r' ;
dt ??^' — 7>li|r' I
similarly | = |/(.) +-0)| (j-^^' - «f ) - |/((),
Multiplying these equations by ?, wi, w respectively, and
adding, remembering that 1^ + m" + w" = ^'^j
,^M c7m f?M , du .
^ J- + ?w -J- + w -J- + « -^ = 0,
GKC dy dz dt '
du „,, ,
smce -J means/ [t).
1850.
1. A paraboloid of revolution with its axis vertical contains
a quantity of water, into which is sunk a heavy sphere, and the
water is just sufficient to cover the sphere ; find the form of the
paraboloid that the quantity of water with which this can bo
done may be the least possible.
Let a be the radius of the sphere, I the latus-rectum of the
paraboloid; h the height to which the water rises when the
sphere is siuik : then if C be the content of the paraboloid of
height ^, V the volume of the sphere, Q the quantity of water,
g = o-r;
and we have to make Q a minimum by the variation of
I and h.
176 SOLUTIONS OF SENATE-HOUSE PK0BLEM8. [1850.
Now C=l'rrlh',
therefore IK^ must be a minimum.
Now from the vertex the equation to the section of the
paraboloid is
/ = ^x ;
that to the section of the sphere is
[x- {h-a)Y + y' = d\
In order that these may touch one another, we must have
[x — {k — a)Y + Ix — a\
a perfect square, which requires that
4:{h'-2al) = {l-2{h-a)]%
or r - U[h-a) + 4a' = 0 (l).
Hence we must make JK'' a minimum subject to the con-
dition (1), which may be written
£+_2ar.
4Z '
therefore we have to make
(Z+2aV
= mmmium
I
4 1
or ^ T = 0 ;
.-. ? = fa;
which determines the form of the paraboloid.
2. If a circle be described touching a curve at any point
(r, 6) and passing through the pole, shew that the equation to
the circle will be
do r
The general equation to a circle passing through the pole
and the point (r, 6) is
r — r sec(^— a) cos(^' — a) (1),
1850.] DIFFERENTIAL CALCULUS. 177
a being the angular coordinate of the diameter through the pole.
This equation may be put in the form
cos{6'-a + (^'-^)|
j« ^= )• '■ !^ ~
cos [d — a)
= r {cos {& -6) - tan {6 - a.) sin {& -6)].
Now from (1), 'ia> = ~ ''' ^^^ (^ ~ '^) '^'^ (^ ~ ^)
= -j^ , when 6' = 6^
do
since the circle touches the curve at the point (?•, 6)\
•'■ *■ r tan (^ - «) = ^ )
and r = r {cos [6' - ^) + "4 sin [6' - 6)]
, d sm[e'-e)
~ '* dd r '
, d sm{0'-0) ^
or r' + r' ^ ^ ' = 0.
dU r
3. If a parabola roll upon a line, the focus will trace out
a catenary.
The following more general problem admits of very easy
solution: A given curve rolls upon a straight line, to find the
locus of any point to which the curve is referred as pole.
Let AB (fig. 79) be the given straight line, A any fixed
point in it. Let CF be the rolling cm've, C the point which has
been in contact with A, S the pole, P the point of contact in the
position represented in the figure. Join ^SP and di'aw /SF per-
pendicular to AB. Let A Y = x, YS = y, SF = r, s the arc
of the cun^e described by S. Then the tangent being manifestly
pei*pendicular to SF, we have
'^ = cohFSY=^,
ds r
* For tills solution, wc are indebted to Mr. Goodwin
N
178 H(.>LUTl(>NS OF SENATE-HOUSE PROBLEMS. [1850.
Let the equation to the rolling curve be r'' =f{p)i then the
equation of the required locus is
^■'{•H- (I)} =/(.).
In the case of the parabola, we have
'■=?^
dx y c
ds r yl
the differential equation to the catenary.
4. Find the forms of the curve whose equation is
,ii I ^i-
xy = m [x + y — « ],
according as nt^ is > = or < -^ a^.
Arranging the equation as a quadratic in cc, we have
and a^ = ^ + A {/ + 4/n' («'-/)}* (1).
Hence we may use x = -^.^ as a guiding cm've ; its fonn
is shewn by the dotted curve.
To consider the equation
or f - 4w* . f + ^m\t' = 0.
This equation, considered as a cubic in y\ will have three
3 3J
real roots or one, according as oti' is > or < — — a'"*.* If it
has three real roots, one of them is negative, and the corre-
3 3-
* If m = -j— «^, it vdH have three real roots, tAvo of them being equal.
1850.J DIFFERENTIAL CALCULUS. 179
spending values of y imaginaiy. If it has two equal, and there-
fore two real roots, the equal roots are positive and the other
negative, giving only one positive value to y'. If it has
only one real root, it is negative and y imaginary. Again,
differentiating the original equation, we get
dy 2m'x — y^
dx ^fx - 2m'y '
therefore
1, when
y
= 0,
and therefore x = a^
dx
Also,
taking
the lower sign in equation (1),
X
=
2m'
2 m \ y^
-1
=
— m
2 2
'"V &c.
= 0 if ?/ = GO :
hence the axis of y is an asymptote to the cui-ve, which we
thus see has the fonn represented in figs. 80, 81, 82, according
3.3-
as m' is > = or < ~ — a^.
4
5. Trace the curve whose equation is
2a
[r — cf = cd {2a — cO) when c = — ,
TT
and prove that as c increases indefinitely the cui've approximates
to a circle.
When 6 is positive, c6 must be less than 2a or 6 loss than
TT, and 6 can receive no negative value. Also, any value 6^ of 0
gives the same value for ?• as tt — 6^. Solving the equation
we have
r = c± [cd[2a-ce)]^
2a
= — + 2a
K(-^)r <■)•
N 2
180 SOLUTIONS OF SENATE-HOUSE PHOBLEMS. [1850.
The quantity affected with the ambiguity lias its greatest
a
value when - = ^, when r receives the value
2a
r = — ± a
IT
the latter of which is negative. Again, putting ^ = 0, we have
r = c] and differentiating,
therefore when r = c, and therefore ^ = 0,
dr
33 = °"-
Hence the cui've is of the form shewn in fig. 83.
Also, as c and therefore a is indefinitely enlarged, equation (1)
becomes >• = — , representing a circle.
TT
6. Find the locus of the consecutive intersections of the curve
whose equation is x^ + y'^ = 2ax' + 2by' (1) ; a and h having
any values which satisfy the equations
2„ (.!-,) = (."-/) I -2., (2),
2i(4-,)=.'-y' + 2.,| (3),
X and y being the coordmates of any given curve.
The problem is best solved by introducing polar coordinates.
Let X = r cos 6^ y = r smO]
.-. dx = — r &mdd6 -+ cosOdr^ dy = r co^6dd + shiddt",
.'. xdy — ydx = r^dd^
[x^ — y^) dy — 2xydx = r^ cos2ddy — r^ sm20dx
= r^ cos Odd — r^ s,m 6 dr^
{x^ —y'^) dx + 2xydy = r^ sin Odd + r^ cos 6 dr.
1851.] DIFFERENTIAL CALCULUS. 181
Hence equations (2) and (3) become
2a = r cos 6 — sin 6 -f^ ,
da
2b = ?• sin 6 + cos 0 -j^ :
au
da . . ^'r
.-. 2^=->-sin^-sm^^,
2 -77: = r cos c? — cos U ^7^ .
Also equation (1) transformed into polar coordinates becomes
r' = '2a cos 0' + 2b sin 0'.
Differentiating this equation with respect to 0, considering
r' and 0' constant, we have
0 = 2 -77. cos^ +2-77= sm^
dtf dv
= r&m{0'-0)-cos{0'-0)'^,-
tan (^'-6') =
dff'
From this equation, when r has been substituted in terms
of 0^ from the known equation to the curve, we can find 0
in tei-ms of 0'', and thence r and -^ will be known in tcnns
dtf
of 0'j and the required equation to the curve will be
r = 2a cos^' + 2b sin^'
dr
= rcos{0'-0) + Hm{0'-0)%.
d0
1851.
If (f) (c) be a rational and integral function of c, the coeffi-
cients of which are functions of any number of variables ^, ?/,,..
then if 8 denote differentiation with respect to the variables,
and the quantity c be eliminated from the equations 0 [c] = 0,
182 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
3(/)(t')=(), the result may be represented by tiSPSQ... = 0
where P, Qy.. are the roots of the equation <f>{c) = 0, and tc = 0
is the result of the elimination of c from the equations (p (c) = 0,
<f>'{c)=0.
Since 0(c) is a rational and integral function of c, and
P, Qy.. are the roots of the equation 0(c) = 0, we have
0 (c) = [c — P){c— Q) identically.
dP dP
Let P^ denote ^- dx. P. -7- diu... then
^ c?^ ' "' dy '^'
g0(c) = 0(c) Slog 0(c)
=-*w(^^^^-^^--)
=-{(^.+^.+-)(o-^)(c-^)-+(^.+^.+..-)(c-^)(«-^)+"l-
Hence the result of the elimination of c between 0(c) = 0
and S0(c) = 0, is
0 = product of the expressions (P^. + P^+...) (P- Q) (P- /?)...,
((?.+ (?,+•••) («-P)(^-P)...,&c.
= (P,+P^+...)(^,+ ^,+...)...(P-(?)(P-P)...(^-P)((2-P)...
= v suppose.
Again, 0'(c) = (c-^) (c-P)...+ (c-P) (c- P). ..+... ;
.-. ,,= (P-^)(P-P)...(^-P)(^-P)...,
and SP=P, + P, +..., 8^= ^,.+ ^^4-...;
.-. v = uBPBQ....
Hence the result of the elimination may be represented by
u8P8Q... =0.
( l«-i )
INTEGRAL CALCULUS.
1848.
The comer of a sheet of paper is tunied down so that the
sum of the edges turned down is constant ; find the equation to
the curve traced out by the vertex of the angle ; find also the
area of the curve.
Let r, ^, be the polar coordinates of the vertex, refeiTed to
the origmal position of the vertex as pole, then the lengths
of the respective edges are
^r sec 0, ^ r cosec ^, respectively ;
therefore the equation to the curve, is
l^?- (sec ^ + cosec ^) = constant = a suppose,
or in rectangular coordinates,
{x + 7/) {x' + f) = 2axy.
To find the area, turn the axes through an angle ^tt, then
we get _acos26'_
*" ~ 2icos^ '
therefore if A be the area of the loop traced out by the vertex,
A=-l rW
1 r\^
i-^ cos' 2^ ,^
27j ff"
COS a
4cos''^-4+-^] iW
cos a/
2cos2^-2 + 6ec'<9)<Z^
= |(2-7r + 2)
184 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
2. Tangents to a system of similar and concentric ellipses
arc drawn at a given pei*pendienlar distance from the centre ;
find the locus of the point of contact, and shew that the area
of the curve is equal to that of an ellipse which has the same
greatest and least diameters.
Take the common axes of the ellipses as coordinate axes,
and let c be the distance of each tangent from the centre, then
if 6 be the inclination of a pei-pendicular on any tangent from
the centre to the axis of a;, the equation to that tangent will be
cccos^ + ymiO = c (1).
Let the equation to any one of the ellipses be
2 'i
7? + !? = '" C^)-
If (^, 7)) be the coordinates of the point where (1) touches
this, we have
f cos^ 7? sin^
' ...n:^ ^ '
and since |, r;, is a point in (2),
m' = K + t'
a: h
Eliminating m^ 6 between these equations, we get
a' "^ h" d' W b"
the equation to the locus of the point of contact.
To find the area of this curve, transform its equation to polar
coordinates. It then becomes
cos^ 6 sin'-^ d
+
a
b'
r = c
cos 6 sin'"^^
and if A be its area, cos'' 0 sin'^ 6
' - -j
A = 2d' ( T-^S '^cW.
cos 6^ sm t^N
1848.] INTEGRAL CALCULUS. 185
Let -7 tan 0 = tan ^ ;
i^ 1 + p tan <^
and A = 2c^ i ' . , - sec^(bdd>
J ^ sec (f) a
= — T I (a'^ sin''' <^ + Z>^ cos"' <^) c?^
Again, let ^' be the area of the ellipse which has the same
greatest and least diameters, then if these diameters be 2rj, 2?\^,
A = Trr/^-
cos'' 0 sin"'' 6
4 "I" 74
JNow r = c
a'
= -5T5 («' cos'''0 + &''' sin'*^) («■'' sin''*^ + h" cos'^0)
cos 20
d'b' (V 2 / V 2
The maximum and minimum values of r will be got by
putting cos20 = 0 and 1 successively;
•*■ "*> ~ a* 2 ' ''•' ~ "■'
., _ TTC^ d' + h'
*'• ~ab ~~2~ '
the same value as that previously got for A.
18<i
SOLUTIONS OF SKNATE-HOUSE PROBLEMS.
[1848.
3. Prove that the remainder after n tenns of the infinite
series — + — + — +.•• whore a > 1 lies between -r^ — -tt^i ■>
f = I — ) and -, w 73-^ , approaehiner much nearer to the
V j^^yx'j (a-l)(« + ir" ^^ ^
former limit when n is large.
(1). In general
1 1
+
^,« + °L(^) ,n-Y +.
[m—pY [m + 2^)
>
Now
+
1
+...+
{n + l-rj))" [n + 1 - {r - 1)2)Y {n + iy
1
+...
+
+
+ l + {r-l]2)Y {n+l+rjjy
+
+ l—rpY {n+l + ypY
+
l{n+l- (r-l)|>] =
+
1
{71 + 1 + {r- 1)2)Y_
1
+...+
{n + iy
> -, r- + 7 rr- +•••+ -, 7T- 1 from above,
[n + \Y [n+lY {n+\Y^ '
2/- + 1
^ [n+lY'
P
+
P
" {n+l-rpY [n+l-[r-l)2iY
Now let f'p = \ - Pi then this becomes
P , P , , i^
+...>
212? + pj
{n + lY
r« +
+...+
{n + i +2)Y [n + ^ + 22)Y {n + i -pT {» + 1)'" '
therefore, a fortiori^
P
+ ...+ T
P
(" + i+7?)^ ■■■ (n + i)* [n + lY
1848.] INTEGRAL CALCULUS. 187
Let j9 = dx^ then this becomes
r"^<& 1
/
>
n.h^'^ (/^ + 1)^'
similarly j^^^^->^^^^^,
>
We shall thus obtain an infinite series of inequalities similar
to the above. Adding them all together, we get
"^ dx 1 1 1
+ 7— T^r, +••■
r dx_
„,_ (a-l)(« + ir {n+iy (n + 2)
Again, if ^j — ^iV being any integer,
+ ^"~ri Tirvr +••• (/ te™is)
(/t + 1 +i^)" (» + 1 + 2^7)^
< -;^ — - — c- + , ^r- +... (;»' terms),
(«+l)" («+l)' ^^ ^'
1
<
(n + 1)^
For /> write dx^ then this becomes
^''+''' ^ 1
/
whence, as above, we get
'" ^ _ 1 1
^ ~ (a-l)(« + l)''' ^ (« + 1)^ ^ (7i + 2)
[ — =
+ 7Z-n7V.+-"5
We have / — - = -, rr ^7 — -Tvsrr - 7 — rsv^l ?
1 r 1
.V 2,1+2) V^^2«+2
2 f a - 1
(a-l)(n+l)'-* (2n4-2
(a-l)(a-2)(a-3) 1
■^ 6 (2n + 2)''^'
1 («-2)(«-3) .
(/?+ 1)' 24(« + i;
188 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
r^ dx 1__ ^ (a-2)(«-3)
R . r ^ - _L_ \ \ L_i
•'„. ^""(«-i) t("+ir (n+2rf'
1 -T TT^U
(a-1) (n+ir i ^1 , 1
n+ 1
a-1 (a-l)(a-2) 1
a-1 (w+ip i(n + l) 2 (n+1
1 a-2 1
(w + 1)' 2 (w+1)^"^
""^•■^^a; a-2
+.
\ r
rx+-
(n + ir J„,, a- 2(,i + i;
It hence appears that when n is large, -. p- approaches
r"+l ^:p r""^^ dx
much more nearly to the limit I — than to I — , whence
the latter part of the proposed theorem readily follows.
4. If f[x) be positive and finite from x = a to x = a + h^
shew how to find the limit of
for n = 00 , and prove that the limit in question is less than
T I f{^) ^^^) assuming that the geometric mean of a finite
number of positive quantities which are not all equal is less than
the arithmetic.
Hence prove that s-'^o'"'* < /„ £'"'*, unless u be constant from
£C = 0 to £C = 1.
Let log/(ir) = -Fix), then
log {/(«)/(« + ', a) .../(« + '^ hj^\
= ?/ suppose.
1849.] INTEGRAL CALCULUS. 189
When n is infinite, let - = dxi then
' n
7/ = ^ {F{a) + F{a + dx) +...+ F{a + h)},
1 /•'
= J I F{a + x) dx ;
...{/W/(., + i ;,).../(„ + ^A)f
approaches to the limit e^-^" iog/t««)''' ^^ ^^f„ iog/(«)rf«^ -^^^ smce
the geometric mean of a finite number of positive quantities
which are not all equal, is less than the arithmetic,
{/(«)/(« + i a). .,/(« + !i^^ a)}*
as long as n is finite, and this will hold up to the limit when n
is indefinitely increased : but in that case
therefore the required limit < y I f[x)dx.
Hence if /(o^) = e", « = 0, and ^ = 1, s-^i""' < /^s"(7a', unless
?i be constant from x = 0 to ic=l, in which case they are
equal.
1849.
1. Investigate the series
&' tt' ^ cos 20 cos 30 „
for values of 0 between — ir and tt.
Let COS0 — ^cos20 + |co830 — ... = n,
sin0 - ^sin20 + ^cos30 - ...= v,
lyO SOLUTIONS OF SKNATE-IJOUSE PROBLEMS. [1849.
then if 0 lie between — tt and tt,
U +-iv = £-*^ - i£-*'^' + ^3
*,
= l0g(l+£-^) = l0g(£ ^U £-') + l0g£-*S
= log(2cosi0) + -H^;
therefore equating imaginary parts,
V = sin0 - ^sin20 + |cos30 -... = ^0:
integrating with respect to 0,
11 R^
- COS0 -f -5 COS20 - -2 COS30 + ... = -+ (7.
To determine the constant, put 0=0;
I 1 ^
Now 1 - |. + ^, -■••= 1 + I + y +••■- 2 (j! + p +•••
II , /. 1 1
. =l+2^. + 3.+-.-Hl + 5. + 3.+.
..osi„e=«|l-(|)]{,-(l)}...,
and sind = 6 — — — ^ +... :
equating coefficients of ^^,
1 /. 1 1
fl'2 TT^ 1 1
and — =T7; — cos0 + ^ cos 20 — -r, cos20 +•...
4 12 2'' S''
2. If a line be drawn through the centre of an ellipse,
cutting the major axis at an angle 0, and the curve at an angle
1849.] INTEGRAL CALCULUS. 191
</), (1) prove that
(a^ - h') cos [20 -</)) = («■' + h') cos</) ;
and (2) that f' (ficie = y .
(1). Let the coordinates of the point where the straight
line meets the ellipse be «cosa, 5 sin a; then will the equation
to the tangent at that pomt be
cos a sin a
X H T- y = 1.
a h "
Hence, by the conditions of the problem,
tant/ = - tana,
a '
T tan0 + tan 6 h
^"*^ A 1 — /] . . = cota.
1 — tany tan 9 a
Hence, eliminating a,
tan'd + tan0 tan0 _ W
1 - tan0 tan</)~~ ~ ~ a"
U - -)i tanfl tan</> = - f ^ + tan'^^ ;
, _ _V cos"' B + a sin' 0
•■• ^^"*P~ "(«•''-//■') sine COS©
^(a-'+^>'-')(cos''g+sin-'0)-(ff''-^/)(cos'''e-siu-'g)
2(«'''-Z>-) sin 0 cos 0
_ fl--' + ^-^ - (^-^ _ ^^) C03 2g
(«■■'- 6^) sin 20 '
.-. (a' - 6') (sin20 tan<^ + cos20) = d' + //,
and («' - />') cos {29 - (f)) = («' + Z»"') cos<^.
(2). Again, since
_ b"" cos' 0 + a" iiW 9
^"^9-- {d'-h-^)^[n9cos9 '
192 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
= / X?n7r - tan -r-r^ — ,2. . ^ ^h at/ (2)
1 C"'
= - I onvdO by adding (1) and (2)
where ?« is a constant integer to be determined.
For this purpose we observe, first, that its vahie is inde-
pendent of any relation between a and h • and secondly, that if
a = b the ellipse becomes a circle and (jy always = ^tt. Hence
in this case
(Pde= ^dB = — and m = - 1.
Hence also, in all cases, m = — 1 and
f
J 0
^dB = --
3. Through a given point B (fig. 84) of the axis of a; a line
is drawn parallel to the axis of y : to any point Q of this line
another straight line is drawn from the origin and produced to
P until PQ = BQ. Find the equation to the locus of P, trace
the cui've, and find the whole area included between the cm've
and the asymptote.
Extend the geometrical description so as to include the whole
of the curve given by the equation.
Let AB = a, AP = r, PAB = 0. Then
QP = r — a sec B,
QB = a tan 0 ;
.*. r = a(sec0 + tan0),
the polar equation to the curve.
1849.] INTEGRAL CALCULUS. 193
DifFcrentiating,
-jji = asccO (t&nO + secO)
da
= r sec 0,
.". r^ -J- = rco&O = a (1 + sinO).
dr
When 0 = - and — , r = oo and r^ -^ = 2a and 0 : hence
2 2' dr '
the axis of ?/, and the line DCD parallel to it such that AC =2a^
are asymptotes to the curve :
f\ r. I dr ^
^ = ^' '• = ^' rTe = ^^
0 > 0, < 2 "■? ^ i^ positive ;
„ TT 1 + sin 0 .
t/ > — < TT, r = — a J. — IS negative ;
2 ' COS0 ^ '
rt Stt 1 - sin 0 .
t/>7r<— -, r = — a jr— is nesrativc :
2 ' COS0 ^ '
B = ^Stt, r = cc ,
„ Stt ^ 1 - sin0 .
0 > — < 27r, r = a ^ - is positive :
2 ' • cost^ ^
the negative values of 0 give no new branch of the curve.
Hence the cun^e is of the form represented in (fig. 85).
To find the area [A) included between the curve and the
as}Tnptote D CD'. Produce AF to meet the asymptote in R ;
then the element of the area A
BA = ^{AE' - AF') 8A = ^pasecOy - a' (sec0 + tan0f} SO
= ^a" (3 sec' 0 - 2 sec 0 tan 0 - tan" 6) BO
= ^d' (2 sec'd - 2 sec0 tan0 + 1) SB ;
.-. A = ^d' f2tan0- -\ + b] + C
' \ cosy /
^ [ \l + smBj j
194 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
from 6 = 0 to 0 = ^TT gives ^A,
.-. ^ = (^7r-f 2) d\
the required area.
If P' be the point where AP cuts the branch BP' of the
curve, it is evident from the tracing of the curve that QP = QB :
hence the curve may be described as the locus of the point P
on the Hne AR whose distance from Q equals QB.
( 195 )
GEOMETRY OF THREE DIMENSIONS.
1848.
1. If three chords be drawn mutually at right angles
through a fixed pomt within a surface of the second order
whose equation is u = 0, shew that 2 ^- will be constant, where
R and r are the two portions into which any one of the chords
drawn through the fixed point is divided by that point.
Prove also that the same will be true, if instead of the fixed
point there be substituted any point in the sm'face whose equa-
tion is w = c.
We shall prove the second part of this only, since it mani-
festly includes the first.
Let the equation to the surface a = 0, referred to its centre
and axes, be
Ax" + Bf + C£' = 1 (1).
Let a, /8, 7 be the point through which the lines are drawn ;
then, since it always lies on the surface u = c, we have
Ad' + ^/3'^ + Ct^ = 1 + c (2).
Let l^m^n^^ ^i^h^t h^^^a'hi ^^ ^^^ direction-cosines of the lines,
then their equations are
-,— = ^ -= - = Px say 3),
h ~ -\ ~ >\ ~^' ^^'
xj-a y - ^ z-y
^'^ =^r=''' <"'
02
106 SOLUTIONS OF SENATE-HorSI-: PROBLEMS. [1848.
('.»'.»,)7 (^.;'».2*'.2)^ ih"',",) lacing, since (3), (4), (5) are at right
angles to one another, subject to the conditions
K' + f: + K =1 (6),
< + < + <=! (7),
< + < + < = 1 (8).
Where (3) meets (1) we have, substituting for xyz in tenns
^ (a + /.p,)'^ + ^ (/3 + m,p^Y + ^ (7 + n.p.Y = L
The roots of this, considered as an equation in p^, are R^r ; hence
Rr ~ Aa' + ^y8^ + Cy' - 1
^ Ai;' + ^m;-^ + c^;^
c
Similar expressions resulting from (4) and (5), we get by
(6), (7), (8),
^ 1 A+B+C
^ Rr~ c '
which is constant.
2. Find the locus of the foot of the pei-pendicular let fall
from the origin on the tangent plane to the surface xyz = a^ ;
point out the general form of the required sm'face, and find the
whole included volmue.
The equation to the tangent plane to the given surface at
a point {xyz), is
^, V, ^,
^ + ^ + - = 3.
xyz
The equations to the pei'pendicular on this plane from the
origin, are
At the intersection of these we have
^^1 = yy^ = ^^1 = - — 3^ ;
therefore, since xyz = a^, we get as the equation to the locus
required, (^^-^ + y^^ + ^-^ = 21a'x^y^z^.
1848.] GEOMETUY OF THREE DIMENSIONS. 197
The form of this surface will be that of four similar sym-
metrical pear-shaped portions, meeting in a point at the origin,
and lying in the octants + + +, -\ , - + — , h-
The equation to the surface, transfonned to polar coordinates,
becomes
r" = 27rt^ cosO slu^0 cos^ sin^.
And if V be the volume of one of the portions
V = ^JfJr' smSdrdddi/,
between proper limits,
9 . /-*"/■-"
= - aM I COS0 sin'0 cos^ sm<f)d6d(f>
2 J 0 -' 0
9 ;
= - a^ j cos (p sin (p dcj)
= ^ a'
16
therefore If V be the whole volume of the surface,
V= 4F' = -a^
4
3. A plane moves so as always to enclose between Itself
and a given surface S a constant volume ; prove that the
envelope of the system of such planes is the same as the locus
of the centres of gravity of the portions of the planes comprised
within S.
Conceive the plane to receive a small twist about any straight
line passing through the centre of gravity of the portu)n com-
prised within S; then, whatever portion is cut off from the
enclosed volume on one side of this line, an equal portion will
be added to it on the other,* so that, by the conditions of the
problem, the plane will pass from any one position to the con-
secutive one by turning about a line passing through the centre
of gravity of the portion comprised within S. Therefore the en-
velope of the planes will be the same as the locus of the centres
of gravity of the portions of the planes comprised within S.
* See Cambridge and Dublin Mathematical Journal, vol. iii. p. 181.
198 SOLUTIONS OF SENATK-IIOUSE PROBLEMS. [1848-
4. OA^ OB^ OC, are three straight Ihies mutually at right
angles, and a lumuious point is placed at C; shew that when
the quantity of light received upon the triangle A OB is con-
stant, the cui"ve which is always touched by AB will be an
hyperbola whose equation referred to the axes OA^ OB^ is
[y—mx) (x — my) = m&^ where OC = c^ and m is a constant
quantity.
With C as centre, and CO as radius, describe a spherical
surface, then the quantity of light received on the triangle
A OB Is the same as that received by the spherical triangle
CA'B' intercepted between the planes COA^ COB^ CBA^ and
will therefore be proportional to the area of that surface. But
if S be this area,
S = ^irr' {A OB' + OAB' + OB' A - it)
= 27rr' ( OAB' -f OB' A - ^tt),
since A OB is a right angle.
Therefore if the quantity of light received by the triangle
be constant, OAB' + OB' A must be so, = 2a suppose.
Let the angle OB' A = a + 0^ then OAB' will = a-d^ and
the equation to the plane ABC will be
cos(a+^) x-\- cos[a — 6)y+ {1 -cos'(a + ^) -cos''(a- ^)}*^=7>;
p will be detennined from the consideration that where this
meets the axis of s, we have s = c ;
.-. {l-co3'''(a+^)-cos'(a-^)li = c,
therefore the equation to AB is
cos {c(.+ 6) X + cos[a— 6) y = (1 — cos'"' {a+ 6] — cos^ (a - 6)]^ c
= (-cos2acos2^)4 c (1).
The quantity (— cos2acos^)* is real, since 2a is greater than
a right angle and less than two right angles, and therefore
cos 2 a negative.
Putting tana = w, tan^ = t, (1) becomes
(1 - 7it) x+ [l + nt] y = {{n' - 1) (1 - f]}^ c,
and we have to find the locus of ultimate intersections of this
line, subject to the variation of f.
1848.] GEOMETRY OF THREE DIMENSIONS. 199
Clearing the equation of radicals and arranging according
to powers of f, it becomes
f [re [x-yf + in' - 1) c'} + 2tn [if-x') + (x + yf - {n' - 1) c' = 0.
Eliminating t between this equation and its derivative, we get
{[x + yY + {n^ - 1) c'] [n' [x - yY + {n^ - 1 ) c'^} = n^ (f - xj,
which may be reduced to
[x + yr-re[x-yf={n^-\)c\
or {(1 -n)x + (1+ n) y} {(1 + n) ic -I- (1 - n) y\ = («' - 1) c' ;
1 f . n — \
therefore puttmg = »i,
[y — mx) [x — my) = hik?
is the equation to the curve always touched by AB.
In a manner similar to this may be solved the following
problem, set in 1851.
Let a spherical surface whose centre is the origin of coor-
dinates meet two of the coordinate planes in the great circles
Zx^ Zy\ also let the points P, Q be taken m Zx^ Zy respec-
tively, so as to make the surface of the spherical triangle PZQ
constant: shew that the curve which is always touched by the
great circle PQ has for its equations x^ -\- y^ -{■ ^ z= d^^ and
xy = ^d^ sin^, where E is the spherical excess of the triangle
PZQ.
The geometrical conditions of this problem are the same as
those of the foregoing. Writing z for c, we have as the equation
to the surface always touched by the plane through the centre,
2 7r + 1 ^ ^ '
+
x' + / + z'
cos2a.
200 SOLUTIONS or Si:NATl->Ht>i:SE PROBLEMS. [1848.
But -rr + E= A' OB' + OAB' + OB' A In the previous notation
= -^TT + 2a ;
.-. 2a = ^TT + E,
and our equation becomes
xy = \ (a;"'* + ^^ + 2:''') sinE".
But since the curve is traced on the sphere, w^e have
x^ -\- y^ + z^ = a^, and we get as the equations to the curve,
X + y '\- z =^ a ,
xy = ^a" sinjE",
the required equations.
5. If 0 be a given point in a surface of the second order,
and OA^ OB, 00, any three chords passing through 0 mutually
at right angles, shew that the plane ABC will always pass
through a fixed point.
Take 0 as origin, and the three lines OA, OB, OC, xn. any
position as axes ; let the equation to the surface be
Ax'->rBy'+Cz'-\-2Ayz+2B'zx-\-2C'xy+'iA"x+W'y+2C"z={).
Then the length of OA will be the value of x, when y = 0,
2 = 0; hence
9 A"
0A = -^-:
A
2B" 2C"
similarly OB = ^- , OC = y^ ,
and the equation to ABC will be
Ax By Cz ^ ^ ,,,
y+^ + ^, + 2 = 0 (1).
And the equations to the normal at 0 are
_ 3/ _
2 A" ~ 2B" ~ 2C"~ 2 [A"" + B"^ + C"''f
where r is the distance from the origin of the point [xyz).
(2),
1849.] GEOMETRY UF THREE DIMENSIONS. 201
Where (I) and (2) intersect, we have, dividing eaeli term
of (1) by the corresponding member of (2),
r
Now we may establish one relation among the nine coeffi-
cients of the equation to the surface. Let then [A"^ + B'"^ + C"'^)
be constant, then the above equation shews that r varies inversely
as ^ + 5 + a
But it is known that if the equation be transformed into the
form
Px' + Qy' + Bz' + 2P"x + 2Q"y + 2B"z = 0,
the quantities P, Q^ B, are the roots of the equation
(-S'- A) [S- B) {8- C) - A" [S-A)- B" [S-B]- C" {S- C)
- 2A'B'C' = 0.
Hence, by the theory of equations,
A+B+C=P+Q + B, a constant.
Hence ?*, the distance from 0 of the point in which the plane
ABC intersects -the normal at 0, is constant, therefore the plane
ABC always passes through a fixed point.
1849.
1. If planes be drawn through any two generating lines
of an hyperboloid which intersect, shew that they will cut the
surface in another pair of generating lines.
Let the equation to the hyperboloid be
'2 2 2
-+^--=1 (1)
Now a plane, drawn through two intersecting generating
lines of an hyperboloid, touches the hyperboloid at their point
of intersection. Let then x\ y\ z\ be the coordinates of this
point, then the equation to the plane will be
XX yy zz
0
+"/-^: = i c^).
202 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
a;', y\ z\ being subject to the condition
or h' c
^+^-:;ii = i (3),
(1) + 2 (2) + (3) gives
m-m'-m- (^)-
a condition which must be satisfied by the coordinates of any
point where (2) intersects (1).
Now (4) may be put into the form
ic + aj'Y'* . _ (z-\-z'^ (y-^y
a
which may be written
shewing that where (2) meets (1) we have either
fi±£:+2=i.fyif: + y±l] and ^^-2 = \ {'-±i - ^f] ,
O/ \C U J Ct rC \ C 0 J
representing one generating line, or
^E±ii + 2 = i' (-£±1 _ 2^) and i^^' - 2 = '., f ^ + ?t±l^)
a \ c 0 J a fc \ c b )
representing another.
Hence if planes be drawn through any two generating lines
of an hyperboloid which intersect, they will cut the surface in
another pair of generating lines.
2. If u =f[x^ y, z) be a rational function of cc, y, 0, and
if w = 0 be the equation to a surface, for a point («, J, c) of
which all the partial diiferential coefficients of u as far as those
of the (w — 1)* order vanish, shew that the conical sm-face whose
equation is
[(x-o) ^ + (3/-^) I + (^-^) j} /(«, h c) = 0,
will touch the proposed surface at the point (a, J, c).
1849.] GEOMETUY OF THREE DIMENSIONS. 203
Tx X — ay — hz — c ,.
Let -^j—=^ = — — 1
/ 111 /rt ^ '
be the equations to any line passing through (a, Z>, c).
Denoting each member of ( I ) by r, we shall obtain the other
points of intersection of (1) with u = 0 by writing
a -f- Ir for cc, h -\- mr for y, c + nr for z
in the equation u = 0. This gives, developing by Taylor's
Theorem,
7 HI 7 HI 7 n«
which, since -^ = -^ = -j-y^^ = 0 for all values of m less than n
becomes, dividinsr out by ,
' ^ ^ 1.2. ..w'
If the line (1) touch the surface ?< = 0 at the point («, J, c)
equation (2) must be satisfied by making r indefinitely small;
(2) will then become
a condition to be satisfied by the direction-cosines of (1) in
order that it may touch m = 0 at the point (a, 5, c). To obtain
the locus of all such lines, we must eliminate /, w, n from the
above equation by means of (1). This gives
j(^-a) I- + (2,-i) I + (.- c) gV{a, J, c) = 0
as the equation to the conical surface which touches u = 0 at
the point (^/, />, c).
204 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
3. A rod AB is fixed to a universal joint to A^ and another
rod BP is conneeted to it by a universal joint at B: all di-
rections of the rod being equally probable, find the chance of
P lying between two spherical surfaces of given radii, whose
connuon centre is A ; and shew that the chance of P lying
within a given elementary portion of space containing the point
P, varies inversely as AP^.
Let AB =a,BP= h.
The chance that P will lie between two spherical surfaces
of given radii, is the chance that the angle ABP will lie be-
tween two values 6^ and 6^, which correspond to the values
r^ and r^ of -4P, r^ and r^ being the radii of the spherical shell.
Now the chance that ABP will lie between 6 and d + B0
= area of zone described by P about B fixed, while 0 has all
values from ^ to 0 + B6 ^ surface of sphere generated
by P about B fixed,
27rb sin^ x b s[n0 , . /, ,/,
4:7rb' ^
Hence the chance required
1 f ^ . ^ 1, z, ■/,^ Ifa' + b'-r;' a' + b'-r
__ I oivi /-/ — I ona H nr\a H \ -— i — i^
- 1 sin ^ = - (cos 0, — cos'^,
2 / „ 2 ^ ' ' ''' 2\ 2ab 2ab
4ab '
The chance that P will lie in an element V of space about P^
= chance of falling in a spherical shell about A as centre of
^, . 1 -, ,. volume of element
thickness or, radius r x ^ ;^-^j — n-
' volume ot shell
_ 2rSr V _ V 1 1
4rtZ' 47rr''*^r STvab ' r r'
4. Determine the condition to which the vertices of a system
of cones which envelope an ellipsoid must be subject, in order
that the centres of the ellipses of contact may be equidistant
from the centre of the ellipsoid.
1840.] GEOMETRY OF THREE DIMENSIONS. 205
Let ^, 7;, ^, be the coordinates of the vertex of any one of
the cones ; then if tlie ellipse be refeiTcd to its centre and
axes, the equation to the plane of contact will be
1:^ + ^ + ^ = 1 (1).
The centre of the ellipse of contact will be the intersection
of (1) with the straight line joining the centre of the ellipse
with the vertex of the cone ; its equations are
X ^ z
Hence if h, A-, I, be the coordinates of the centre of the
ellipse,
In order that the centres of the ellipses may be equidistant
from the centre of the ellipsoid, we must have
Ji^ + I? + T^ = constant, p^ suppose ;
the equation to the locus of the vertices.
5. Determine the form of the termination of a honeycomb
cell on this principle, that if a sphere which will just pass
through the hexagonal transverse section be dropped into the
cell, the unoccupied space at the extremity of the cell shall
be the least possible.*
Let ahe (fig. 86) represent half of one of the rhomboidal
plates, three of which close each hexagonal cell. ABC re-
present the eighth part of a sphere. Then, by the general
principle of Envelopes (see Cambridge and Dublin Mathematical
Journal^ vol. iii. p. 181), the volume in question is least when
the point of contact d is the centre of the rhomboid: or we
must have ac = lad.
Let OA = a, 07) = r ; .'. a = r cos .30° = r - .
* For this solution we are indebted to Mr. Goodwin,
206 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
Let Oac = 0 J .-. a = Od = ad tan0 = ^ac tan^
= ^r cosec^ tan0 ;
. 3* , 1
COS(f> '
.'. COS 9 = ,-T .
3*
Again, hd = ED sin 60 = r — , and ad = ^r cosecc^ ;
.'. t&nbad = ^ = 3* sm(f> = 3* (1 - i)4 = 2*.
These are the angles required.
6. The tangent plane to a surface S cuts an ellipsoid, and
the locus of the vertex of the cone which touches the ellipsoid in
the curve of intersection is another surface S'. Prove that S and
S' are reciprocal, that is, that 8 may be generated from 8'
in the same mamier as ;S^' has been generated from 8.
Take the axes of the ellipsoid as coordinate axes, and let
the equation to 8 be
^ = 0. ° •
That to its tangent plane at any point xt/z, is
, V d8 I , 68 , , d8
This may be put under the form
d8 d8 d8
dS dS d8~
dx "^ dy dz
If ^, 77, ^, be the coordmates of the vertex of the cone touch-
ing the ellipsoid in the cm-ve of intersection with this plane,
we have ^
^ dx
a'^"^ dS dS'
dx ^ dy dz
with similar expressions for t] and ^.
1849.] GEOMETRY OF THREE DIMENSIONS. 207
Again, if with [xyz) a point of S as vertex we describe a
cone touching the ellipsoid, the equation to the plane of contact
will be
«l , ^ , ^ _ 1
d' ^ h' "^ 6' ~ '
1^, 77, f being its current coordinates. To find the locus of
ultimate intersection of these planes, eliminate .r, ?/, z between
the differential of the preceding equation, and of
this gives ^^ + ^ = 0,
h dy
Multiplying these equations in order by a?, y, 2, and adding,
^'eget js dS dS ^
^ + ^;^ + ^^ + '^=''
dS
^ dx
dx " dy dz
with similar expressions for 77, ^.
Hence the locus of f , 97, ^ is 5".
That is, the locus of the vertex of the cone touching the
ellipsoid in its curve of intersection with a tangent plane to 8
is the same as the envelope of the plane of contact when a cone
is drawn from a point of 8 as vertex, circumscribing the
ellipsoid. This holds for all surfaces, therefore for 8' .
But from the mode of generation of 8\ it is easy to see that
the envelope of the planes of contact of cones drawn from its
points as vertices is 8\ therefore, by what has been proved, the
locus of the vertices of the cones touching the ellipsoid in its
curves of intersection with the tangent planes to 8' is 8^ that
208 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
is, S may be generated from S' in the same manner as S' was
from S, or S and S' are reciprocal.
1850.
1. If from a point 0 be drawn any two lines to the polar
plane of 0 in a surface of the second order and meet the plane
in A and i?, and if the central conjugate plane of OA meet OB
in C, and the central conjugate plane of OB meet OA in Z),
CB is parallel to AB.
Take that diameter of the ellipsoid which passes through 0,
and two diameters conjugate to it, as axes. Let the equation
to the ellipsoid be
H 2 2
X y z
1- — -I = 1
a b c
Let ^ be the distance of 0 from the centre, then the equation
to its polar plane is
a
" = T-
Let the coordinates of ^ be — , ?/j s^ ; of ^, -^ , y^, z^. Then
the equations to AB are
a'" y — V, z — z^ ,.
f 3/l - 3^2 ^1 - ^2
The equations to OA will be
a"
X ^
t y z
:75 ^ = f" = ^ = ^ suppose (2),
I .
therefore the equation to its central conjugate plane is
'a"^ J\ X w,?/ z^z
l-f)5=' + ¥ + iP = ° <-'>■
Similarly, the equations to OB are
'2
a
^ — -y
^^ = ^ = i = .-, (4),
1850.] GEOMETRY OF THREE DIMENSIONS. 209
therefore that to its central conjugate plane is
At the point C, the intersection of (3) with (4), we have
1-^
Similarly, it will be seen that at i), the intersection of (2) and
(5), we have
a
'2 / '•! ,
the values of .r at C and D are each equal ^o -g + (^ - I) »',?
therefore the equations to CD are
«^ = T + h--^ri'
I VI J '' y,-y^ ^1 - ^2 '
by comparing which equations with (1) we see that CD is parallel
to^i?.
2. A plane moves so as always to cut off from an ellipsoid
the same volume ; shew that it will in every position touch a
similar and concentric ellipsoid.
If a plane be drawn touching the interior of two similar
and concentric ellipsoids, the point of contact will be the centre
of its elliptic section made by the exterior one. Now conceive
this plane to receive a small twist about any diameter : it will
still remain in contact with the interior ellipsoid, and whatever
portion is taken from the volmne intercepted between it and
the exterior ellipsoid on one side, will be added to it on the
other, therefore that volume will be unaltered. Hence con-
versely, it follows that if a plane move so as always to cut off
p
210 SOLUTIONS OF SENATF.-IIOUSE PROBLEMS. [I8o0.
from an ellipsoid the same volume, the sm*tacc which it always
touches will be a similar and concentric ellipsoid.
3. If F{x^ y^ c) = 0 be the equation of a system of curves,
where c is a variable parameter, and (f) (.r, y) = 0 the equation
of the envelope of the system ; shew that ^ (.x, y) = 0 is the
equation of a cylmder whose intersection with the surface
jP(.r, y, s) = 0 is the locus of points which in sections parallel
to the planes of yx^ zx^ have their tangents parallel to the
axis of z.
Ex. The cone whose equation is a? + y'^ + s^ = [Ix + my + nzj
is cut by planes parallel to the planes of yz and zx ; find the loci
of the extremities of the diameters of the sections which are con-
jugate to the vertical diameter.
(a). The equation (f>{x,y)=0 (1)
results from the elimination of c between the equations
F{x,y,c) = 0,
and -^- = 0.
dc
It will therefore be also obtained by eliminating s between
F{x,y,z)=^0 (2),
and -^ = 0.
dz
Hence, where the sm-faces represented by (1) and (2) inter-
sect, we have
f- (3).
Now the equation to a tangent plane to (1), parallel to the
plane of yz. is
, ,dF , ,dF ^
1850.] GEOMETRY OF THREE DIMENSIONS. 211
dF
If tlicrcforc -^- = o, this becomes
az
which is evidently parallel to the axis of z.
Hence at the curve of intersection of the cylinder (1; and
the surface (2), the tangent in a section parallel to the jdane
of yz is parallel to the axis of z.
Similarly it may be shewn that the tangent in a section
parallel to the plane of xz is parallel to the axis of z.
(/3). In the example, the tangent at the required points are
parallel to the vertical diameter, tliat is to the axis of z^ hence
we get the locus required by eliminating z between
F{x^ y, z) = oc' + / 4- z' - [Ix + my + nzf = 0,
dF
and -T- = 2z — 2n [Ix + my + nz) = 0.
The latter equation gives
Ix + my
1 - n^
Hence a.'" +f= [Jx + my; (l + ,^^ - n' ^^^^1'
_ {Ix + myY
is the equation to a cylinder, whose intersection with the given
surface is the required locus.
4. If .r, ?/, 2-, be the coordinates of any point P on the surface
f{x, y, z) = 0, x\ y\ z of a point F on the surface /(a?', y\ z') = 0,
and for any position of P, x' = Ix^ y = my^ z = nz ; and if the
surfaces be such that when we take any two points P, Q on the
first and two corresponding points P', Q' on the second, PQ is
equal to P ^ ; find the fonn of the sm^aces.
Let f , 17, ^ be the coordinates of Q.
Then ?f , W17, nf are those of Q\
... PQ- ={x- I^Y +{y- mriY + [z - nXf,
PC/ = (I - Ixf + (7; - my? + (^-. nz)%
P2
212 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
and these are equal ; hence we get
(^ - I^y + {y- mvT +{z- n^Y = (^ - Ixf + (77 - m^/f + {^- nz)\
oi'{l-r')x'+{l-ni')f+{\-7{')z'={\-r')^'+{l-m')v' + {l-n')^\
for all values of x, 3/, z ; |, ■»;, ^, consistent with the equation to
the surface /(.r, y, z) = 0.
We must therefore have
(1 - r) x' + (1 - m') y' + (1 - '?'") z' = constant, a' suppose,
which determines the form of the required surfaces, which are
evidently central smfaces of the second order, of which the axes
of coordinates are principal axes.
5. It is not possible to fill any given space with a number
of regular polyhedrons of the same kind except cubes, but this
may be done by means of tetrahedrons and octahedrons which
have equal faces, by using twice as many of the fonner as of
the latter.
Consider two octahedi'a so placed that two of their edges
shall coincide, and the squares of which they are sides be in
the same plane. Let AB (fig. 87) be either of these edges,
G a vertex of one octahedi'on, not lying in the plane of the
squares, D the corresponding vertex of the other. Then CD = a
side of the square = AB = CA = CB = BD = AD^ by definition
of a regular octahedron. Hence CADB is a regular tetrahedi'on.
Hence if we have a number of octahedra, so placed that one
plane shall contain a square section of each, and each edge of
each such section coincide with one edge of each of the adjacent
sections, an equal number of tetrahedra will fill up the vacant
space above the plane, and therefore by using twice as many
tetraliedi-a as octahedi-a, we fill up the space above and below.
6. Prove that the tangent plane at any point of the surface
[axY + iJyyY + [czY = 2 {bcyz + cazx + ahxy)^
intersects the surface ayz + hzx + cxy — 0 in two straight lines
at right angles to one another.
1850.] GEOMETRY OF THREE DIMENSIONS. 21.3
The equation
{axY + [hjY + {cz)'' = 2 [hcyz + cazx + ahxy)
may be put into the fomi
{ax)^ + [hyY- + [czY- = 0 (1).
Also ayz + hzx + cxy = 0 may be wiitten
« ^ c „ ,_.
- + -4 - = 0 2 .
x y z
The equation to the tangent plane to (1) at any point [xyz] is
Let [l^m^n^^ iO^h^^) ^® ^^^ direction-cosines of the lines m
which (2) meets (3), then the condition of these being at right
angles to one another, is
IJ^ + m^m.^ + n^n^ = 0 (4).
Now where (2) meets (3), we have, writing x^y^z^ for xyz in (2),
«(ir-(r-(ir-(-)'i(iy^(i)'a'
a quadratic m -- whose roots are — ^ , — ^ . Hence
©'^.-(D'^'-e/^-" (^'-
njWjj VC2;
similarly -^-^ = f — ) :
n^n^ \czj '
.-. IJ^ + ?n,?«, + »,?i, =c (aa-)* + (%)* + [czf = 0 by (1).
Hence the tangent plane at any point of (1) cuts (3) in two
straight lines at right angles to one another.
7. A certain territory is bomided by two meridian circles,
and by two parallels of latitude which differ in longitude and
latitude respectively by one degree, and is known to lio within
certain limits of latitude : find the probable supci*ficial area.
214 SOLUTIONS OF SENATE-HOUSE TROBLEMS. [1851.
First, to find tlic ohaiiec that tlic centre of the territory
which lies between known limits a, /9 of latitude lies in the zone
between paraHels of latitude J and I + hi.
area of zone breadth hi
This chance =
area of zone breadth (a— /3
27r?-cos/ rhl
I 'Inrr cos^ rhl
cosZ hi
(/• the radius of the earth,)
sin a — sin /3 '
Then the probable supei-ficies of the territory
co%lhl
j li sina — sin/S '
A being the area of the territory when its centre lies in the zone
between the parallels I and I + hi;
1 W+30
= — - {sin (l + 30') - sin [l - 30')},
180
27rr'''sin30'
cos/.
180
Therefore the probable superficies,
27rr'sin30'
/3 180(sina — sm/3)
7rr'sin30'
cos' Ihl,
180(sina-sin/3)
[a - y8 + 1 (sin2a - sin2^)}.
1851.
1. A line passing through a fixed point and having the sum
of its inclinations to tAVO fixed lines through the same point
constant, generates a cone of the second order.
Any section perpendicular to either of the fixed lines has for
a focus its intersection with the fixed line.
1851.] GEOMETRY OF THREE DIMENSIONS. 215
(a). Let OP be the moving line, OS^ OH the fixed lines.
Take the lines bisecting the angle 80H and its interior angle
respectively, as axes of x and y. Describe a spherical surface
about 0, cutting OP, 08, OH in P, 8, H-, join SH, SP, HP
hy arcs of great circles, then by the conditions of the problem
SP + HP = constant, 2a suppose.
Bisect ^7/ in X, join OX, let SX = HX = ^. Draw PM
an arc of a great circle, perpendicular to SH, let XM — 0,
PM= (f). Then, by Napier's rules
cos SP = cos 8M.C0S MP,
= C0S(/S+ d) COS(j>,
COS HP = cos HM.cos MP,
= cos(/S— 6) coacf).
XT OT> TTT> o SP+HP SP-HP
JNow cos oP + coaiiP = 2 cos - — cos ;
.'. cosp cost/ cos(^ = cos a cos ,
. jrj, Qp o • SP'rHP . SP-HP
and cos HP — cos bP = 2 sm sm ;
' a ' a M ' ' 8P-HP
.'. smp sma cos 9 = snia sm ,
therefore adding squares
,, , fcoB^ 13 cos' 0 sm'8mr^0\ , ,
cos'</) ^^— + ~, = 1 (1 .
^ \ cos' a sm'a j ^ '
Now sln^ = sinPil/ =
x'-\-y'-\-zy
2 » ^ + y
.-. cos 0 = —, ^ ; ,
^ x^ + y^ + z'^
cos 0 = j-z 2^ 7 i^i" 0 = / ., ,M ;
{x'+yy {^' + y')^'
therefore equation (1) becomes
x' — ^ + y ^^ = x' + y' + z' 2),
cos'a "^ sm'oL ^ ^ '^
shewing that tlie locus <»f /'is a cone of the second order.
216 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
{j3). Let i'cosyS + ?ysiii/S = ;; (3),
be the equation of a plane perpendicular to OH.
Where this meets OH, we have
X = p cos^, y — V sinyS.
The distance of any point [xyz] in (3) from this point is
If the point {xyz) also lie in (2), this becomes
fx' cos'/3 ^ ?/ sin'^/3 _ ,\4
V cos"''a ' sin'''a / '
or substituting for y from (3)
jiC'^ COs'-'/S (j3-£CC0S/9f 2I*
( cos'^a sin"''a J '
which is equal to
x' cos'' /3 'Ijix cos /3 ^/ cos^ a\ *
cos a sm a sm a sm a
a; cos/3 cosa
or ; p — — .
cosa sma sma
Hence the distance of any point in the curve of intersection
of (2) and (3) from the point of intersection of OH with (3) is
a linear function of x, which is a property peculiar to the focus.
Therefore any section pei^pendicular to either of the fixed lines
has for a focus its intersection with the fixed line.*
2. The locus of the points in which a prmcipal plane of a
surface of the second order is intersected by the noi-mals at the
different points of a plane section of the sm'face is a conic
section.
Let the equation to the surface referred to its prmcipal
planes, be j^-^^ ^ Bf + C'/ = 1 (1),
and to the plane of section
Ix + lay + nz =2^ (2)-
* Sec Ilcaru on Cunes of the Second Order, p. 60, ct scqq,
1851.] GEOMETRY OF THREE DIMENSIONS. 217
Then those to the normal at [xyz] are
Ax By Cz
Where this meets the plane of yz^ we have
i',=3'(i-D, ^.=^(1-2
Ay, Az,
EUmhiathig x between (1) and (2), and substituting the
above values of y and z^ we shall obtain the equation to the
locus required, which may easily be seen to be of the second
order.
3. Noraaals are drawn to a surface at points indefinitely
near to and equidistant from a fixed point in the surface : de-
temiine and discuss the equation of the surface generated by
the nomials.
Take the fixed point as origin, and the principal planes
through it as planes of yz and zx. Let the equation to the
surface be ^ = ^^.2 ^ ^y. ^^^^^
The equations to the noraials at (a*, y^ s), are
or x^- X ->r 2Ax{z^-Ax'- By'-...) = 0 (1),
y,-y + 2By{z^-Ax'-Bf-...) = 0 (2).
Again, since the point (xyz) is always at the same distance
fi'om the origin, we have
X + y + z = rt ,
or x' + y'+ {Ax' + Bf+...Y = a' (3).
The elimination of x, y between (1), (2), (3), would give the
equation to the surface. But since the point [xyz] is always
indefinitely near to the origin, .r, y, z arc always indefinitely
218 SOLUTIONS OF SENxVTE-HOUSE PROBLEMS. [1851.
small, and \vc may neglect their powers higher thau the second.
Hence our equations become
.-r, — X + 2Az^x = 0,
y,-y + 2i?2!,7/ = 0,
Eliminating a?, y between these, we get
< + y^ - .2
(2^2, -1)"^ ' i^lBz^-X)' '
the required equation to the surface.
This surface is evidently of the fourth order, and symmetrical
with respect to the planes of yz and zx. Its section, by any
plane parallel to the plane of xy is an ellipse, which becomes
a circle when the distance z^ of the cutting plane from that of
xn = ^ . When z, = — r , the equation becomes x, = 0,
"^ A + B ^ 2>1 ' '
shewing that the section is there a straight line parallel to the
axis of y, and similarly when ^1 = ^5 the section is a straight
line parallel to the axis of x. The points where these lines
meet the axis of s, are the centres of curvature of the principal
sections for —7 , -^ are the principal radii of curvature at the
origin. When ^1 > ^ (supposing A > i5), the area of the
section continually increases, as manifestly ought to be the case,
since the normals altogether diverge after ^1 > ^ •
4. A plane is drawn through the axis of 3/, such that its trace
upon the plane of zx touches the two circles in which the plane
of zx meets the surface generated by the revolution romid the
axis of z of the circle [x — aY + z^ = c^ (c < a) ; find the equation
to the curve of intersection of the plane and surface, and from
this equation trace the curve.
1851.] GEOMETRY OF THREE DIMENSIONS. 219
The equation to the phinc will bo
z X
(0-
c [d'-c^f
The equation to tlie surface, generated by the revolution
round the axis of z of the given circle, is
\{x' + yy--aY^z' = 6\
which rationalized becomes
[x' + if + z'-^d'-6y = lc{'{x'^y'') (2).
To obtain the curve of intersection of (1) and (2), we must
turn the planes of ^^ and xy round the axis of _y till (1) coincides
with the plane of xy^ and then put z = 0. This is effected by
^^'I'iting [a^_(^\i.j,-c.^
tor X.
a
ex + (a^ — c^)^ z .
^ — for z.
a
or since z is to be put = 0, ^^ — x for x. and — for z : this
reduces the equation to
(a;« + f + a' - 6y = 4 [{d' - &) x' + a'y },
or [x' + y' + d' - c'Y = -i{a'- c') {x' + /) + 4o\y'-',
which may be reduced to
x' + y' - a^ + c^ = ± 2cy,
or x' + [y± c)' = d%
shewing that the curve is composed of two circles, the radius of
of each of which is a, and whose centres lie on the axis of y, on
opposite sides of the origin, and at a distance from it = c.
5. Prove (one of) the two following properties :
(1). If [A]^ {D) be two given spheres not intersecting each
other, then every sphere which cuts [A] and (/>) in given angles
will touch two fixed spheres.
220 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
(2). If (/I), [B] be two given spheres cutting one another,
then every sphere which cuts {A) and {B) in given angles will
cut orthogonally a fixed sphere.
(I). Take the line joining the centres of the spheres for axis
of .c, and its middle point for origin : let a, — a be the abscissae
of these centres, r,, r^ the radii of [A) and [B] ; x, i/j z the
coordinates of the centre of a sphere which cuts {A) and {B) in
given angles a, /S ; r its radius. Then we must have
[x — ay + / + z^ = r'^ + 1^ — 2rjr cos a,
[x + (if + if + z' = r^l; + r' - 2?y cosyS.
Let (&, 0, 0) be the coordinates of the centre of a sphere
which this moveable sphere always touches.
Adding and subtracting the above equations, we find
ic' + a' + f + z' = \[r'^ + r/) + r' - v{i,\ cosa + ?'._, cosyS),
and — Ixb = - - \rf — r^ — 2r [r^ cosa — r^ cos/3)],
y^ - d' =^F - a\
adding these three equations, we have
{x —ljf-\- f + ^^ = ?•' — r\ r^cos a + )\ cos /3 + - [ii\ cos a — r, cos /3) \ ,
+ i{<^ + ^-.^ +^ (^-/-O} + ^"^ - «^ (!)•
It is evident that the moveable sphere will touch the sphere
whose centre is at a distance from the origin if the right hand
member of equation (1) be a perfect square, or if
jr^cosa + r^cosyS + - (r^cosa— r^cos^S) I = 2 |rj^+ r./+ - [rf— r./) \
a quadi*atic for the detennination of J, shewing that there are
two spheres which the moveable sphere always touches.
(2). Also it is evident that the moveable sphere will always
cut orthogonally the sphere, the abscissa of whose centre is Z»,
if the right-hand member of equation (1) assume the form
1851.] GEOMETRY OF THREE DIMENSIONS. 221
This it will do if
r^ cosa + r^ cosyS + - {i\ cosa — r^ cos/3) = 0 ;
h »• COS/8 - r, cosa
or — = ~ ' '
a r^ cosyS — r^cosa '
which dctennines the centre of the sphere.
( 222
DIFFKIIENTIAT. EQUATIONS.
1848.
cos 37
Assuming that sinx H is a particular integral of the
equation
S + (-l)^;« (■).
find the complete integral of the equation
S-(-l>=^" (^)-
We see by substitution that not only
cos a;
y = smr» H ,
^ X '
is a particular integral of equation (1), but also
since
^ X
Hence the complete solution of (2) is
. / . cos£c\ -r, ( sin a?
y =■ A\ sma; H — j + i? I cosa-
where A and B are arbitrary constants.
Now assume as the integral of equation (2),
. / . cosa-N „ / sin.r\
y ~ A\ sma; ^ J + i> ( cosa; 1 ,
where A and B are now functions of x which have to be
detennined.
By the usual assumptions of the method of variable para-
meters, we find
dy . f sina; cosa?\ -r. / . cosa^ sina;\
-^ = A\ cosa; s— ] - B [ sma; H -^- ,
ax \ X X j \ X X J ^
, dA I . cosa;\ dB ( sina;\ ^ ,^.
and T- sma; H + -^ cosa; = 0 (3).
ax \ X J ax \ X J ^ '
1849.] DIFFERENTIAL EQUATIONS. 223
., dA ( sin a? cos.rN dB ( . cosa; 8111,0?
Also ^— cosa:: 5- ?- sin^c -\ ;
ax \ X X j ax \ x x
= ^' (4).
From the equations (3) and (4), wc proceed to find -y- , -7- ,
dA {[ sinicV'' cos a; / siiiicN / . cosoj'
-5— \ cosic ^ cosa? 4- sma? ^
ax y\ X j X \ X J \ x
sin a; / . cosicXl
- -w r"^ + — j[ •
2 f sina;
= X cosa;
V a;
dA [^ 1 1\ „ / sina;\
or -7- 14—, 5 = a? cosa; ,
dx \ X' X I \ £C / '
dA „ / sina;\
, „ ,„, dB „ / . cosa!\
and irom (3) — =. — x \ sina; 4 j ;
.'. A = a^'^'sinx — 3/a; sina%?a;,
= a;^ sina? 4- 3a; cosa; — 3 sina? 4- C,
and B = x^ cosa? — 3/a; cosa;^,
= x^ cosa; — 3a; sina; — 3 cosa? 4- D ;
therefore the complete integral of (2) is
/ . cosa;\ , „ . ^ „ . ^.
.•. y = [ sma? 4- 1 (a? sma; 4- 3a; cosa? — 3 sina? 4- C j,
/ sina?\ ... „ , „ ^,
4- ( cosa? ] (a? cosa? — 3a? sma? — 3 cosa? 4- i/j,
2 ^J . cos.-z;\ ^ / sina;^
= a;^ 4- 6 [ sma? H \ -^ D{ cosa? j ,
C and D being arbitrary constants.
1849.
A curve is defined by this property, that the radius of cur-
vature at any point in a given multiple of the portion of the
224 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
nonnal intercepted between tlic point and tlie axis of abscissa? ;
prove that the length of any portion of tlie curve may be ex-
pressed in finite terais of the ordinates of its extremities.
The lengths of the radius of curvature and normal are
respectively
dx)
hence the dilFerential equation to the curve is
^"^'^LL = „Ji^r^V
= ny \ 1 -I-
d'^y "^ \ \dxj
_ A
dx"" 1
(Fx
dy' 1
or y I = — .
Lt — -t, B' .^-Ii (^^W —
dy - ^^^ ' " dy ~\ [dyj ] dy '
and cot^ -7- = — :
dy ny '
.". log cos^ = ^ogCy ;
or cos 0 =
(GuY
n
and s = -^ C^y " + 0',
C, C being arbitrary constants.
1850.J DIFFERENTIAL EQUATIONS. 225
lleucc the lengtli of any portion of tlie curve is known in
terms of the ordinates of its extremities.
1850.
1. lff{x — a, y — h^ z — c) be homogeneous with respect to
X — a^ y — h^ z — Cj then y (a; — a, y — b, z — c) =0 is the equa-
tion of a cone whose vertex is (a, J, c) ; if the cone pass into
a cylinder by «, J, c becoming infinite, shew algebraically that
the limiting fonn of the above equation is
(j) [nix + ny -^-pz -j- q^ mx -f ny +p'z + q) = 0.
Let the axis of the cylinder, to which, as its limiting form,
the cone tends as a, Z», c, are indefinitely increased, be parallel
to the intersection of the planes
mx + ny + pz = 0,
m'x + n'y + p'z = 0.
Then we have ma + nb + pc = a finite quantity, a suppose,
m'n -{ 71 b + pic = a'
Hence when «, Z>, c, become infinite we get, neglecting a, a!
in comparison with a, b^ c,
-.^^=^^ = -.^-^ 0).
np — np pm — pm mil — mn
Now since f is a homogeneous fimction, we have, if n be
its degree,
= 0,
if iV if if if if ,„>
Hence dividing each term of the left-hand member of (2)
by the coiTcsponding member of (1), and observing that when
a, 6, c, become infinite, the right-hand member will vanish after
the division,
i¥ - "» ^ + (i^»'' -2>"i) ^ + (»'i«' - »«'«) '£ = 0...(3) :
Q
226 SOLUTIONS OF SENATE-HOUSE PKOBLEMS. [1850.
whence, by Lagrange's method, we get
dx dy dz
np — i^p pm — p'm mn — m'n '
whence mdx + ndy + J)*^^ — ^j
m'dx + ndy + p'dz = 0 ;
.". mx + vy -^ pz -\- q =0,
m'x 4 ny + p'z + 5'' = 0,
q^ q being constants.
Therefore the integral of (3) is
(f> [mx + ny + p^ + q, m'x + n'y + p'z •+ q) = 0,
the limiting fonn of f{x — a, y — h^ 2; — c) = 0, when a, />, 0,
are indefinitely increased.
2. Prove the following formulae :
W- - = '«(^ + ^') (1X5:7 + 93113X5 +•■•)•
cosl(,>-2.-) «-.l ^ cs{{n-2r)b-s] ^ ^^^^^^
^ ' sm(a— c>)sm(a— cj... sm(6— a) sm(o— c) . . . '
where s is the smn of the n quantities a, 5, c,... and r is any
integer between 1 and n — \ inclusive.
(1). We have
Vl.3.5.7 "^9.11.13.15 "^*"
= ^{(l?7"3:5) + (9j[5"rL13)+-
= {(!-}) - 3 (^-i)} + {{h-h) - 3 (,^1-1^)1 +•••
= 2(j-H^-U..-)-(l + ^-^-|+-) (!)•
1850.] DIFFERENTIAL EQUATIONS. 227
Again, generally
1 x' X* x'' 1 , n+x
l+J + -5+T'--=2-x^'^[l-
Let ic* = — 1, then x = cos|7r H — ^sin^Tr,
1 + ic _ 1 + cos^TT H — * sin^TT
' ' 1 — X 1 — COS^TT - — 4 sin^TT
2 cos'''|7r H — * 2 sin^TT cos^tt
2 sin^^TT * 2 sin^TT cos^tt
, 1 COSiTT -\ * siniTT
■ = COtiTT . f J 1-
Sin^TT * COS ^77
= — * COt^TT
= £-*^"coti7r;
^^^ (r=^) = - H-^ + log COt^TT,
and
2x 2* + - 4 2*
Oj i 9*
, /l+a;\ 2* 42* , ,
; ^^Ir^j = 4 (- ^TT + log COt^TT)
= ^i (log COt^TT + ^TT) - -i — ( log cot '^ - ^TT
Therefore, equating real and imaginary parts,
1 -
-l + l
i
2,2* (^^^ ^^■^^'^ +
i^),
1
3 ~
HA
-
... =
2.2* (^"S
COt^TT
-i^);
1 -J- '
1 1
57
+
... =
TT
2^4'
AH (
1
+..
•)=
TT TT
2 2.2*
' ^^ I1.3.5.7
TT
2 (2 + 2*) '
7r = 96 (2 + 2*) f— L- + ^ +.
^ ^ Vl.3.5.7 9.11.13.15
228 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
(2). In general,
Let y = wx, 0) being one of the imaginary cube roots of unity,
then £"" = ^ + -]- "^TT "^■"
Similarly, e'"'^ = l + ^ + ^V..,
^^=^+f+i4+-
Now 1 + ft) + ft)^ = 0,
But e'"" + £"'•'■ = e^-^^-** )^ + e(-5--*i )^
= £ - 2 cos --
2 y '
1.2.3 1.2..3.4.5.6 ^ V "r -r ;
= ^('£" + 2£-^-'cOS^
This problem may also be solved by putting the series = w,
we shall then get the differential equation -^ — t« = 0, the in-
tegration of which, when the arbitrary constants are properly
determined, will give the required value of u.
(3). If r lie between 1 and r?, we may assume
cos (?i + 1 — 2r) ;r _ A B
sin(ic— a) sin(ic — 6)... sin (a? — a) sin(a; — J) "*'
^, J5, ... being quantities independent of iP,
.•. cos(»i+l— 2r)a;=-<4sin(ir— J)sin(a;— c)...+jBsin(a7— a)sin(a;— c)...
+ . . . identically.*
* We may justify the above assumption by expanding both sides of this
equation in terms of sinx and cosx, and dividing by cos"''j; ; the left-hand side
1851.] DIFFERENTIAL EQUATlOiNS. 229
Putting X = rt, we get
cos {n+ l — 2r) a = A ain {a — b) sin (a — c) . . .,
^ cos(n+ 1 — 2r) a , >
8in(« — oj sin(a — cj...
t3- -1 1 7? cos(w+l-2r)6 , .
bmiilarly B= • /, x • /z.-^ — (2 ,
•^ 8in(6-a) 8m(6 — cj...
cos (« + 1 — 2r) a? cos (n + 1 — 2r) a
sin(ic — a) 8in(cc — 5)... 8in(a — J) 8in(a — c)...sin(a; — a)
cos (n + 1 — 2r) 5
sin [b — a) sin (& — c) . . .sin [x — 5)
cos (n + 1 — 2r) a cos (n + 1 — 2r) &
'* sin(a— &)8in(a— c)...sin(a-a;) 8in(5— a)sin(5— c)...sin(6— a;)
+ . ^°«(''+!--2'-)-^ ,0 (8).
8in (aj — «) s,in[x — 0) . . .
In a similar manner it may be shewn that
sin [n + 1 — 27-) a sin [n + l — 2r) ^
sin(a— &)sin(a— c)...sin(rt— x) sin(5— a) 8in(5— c)...sin(6— ic)
, _ sin (m + 1 - 2r) a; _^
sin(a; — a) sin(a; — J)...
(3) coss + (4) sins, where s = a + b +...+ x gives
cos{(n + 1 - 2r) a- s} cos{(«+ 1 — 2r) & — s} _
sm [a — 6) sin (a — c) . . . sin {b — a) sin [b — c)... '" ^ '
the required result proved for the w + 1 quantities «, J,... a^.
of the equation becomes /(tanx).(seca:)^''''", /being of n — 2r + 1 dimensions,
or /(tanx) (1 + tan*^;)'''', which is therefore of n — 1 dimensions, and the
equation becomes one of n — I dimensions in tanx ; it may therefore be
identically satisfied by the n quantities A, B . . . . We here suppose » + 1 — 2r
positive: if however 2r > n + 1, we may A\Tite for 2r, 2m + 2 — 2s, where
2a > 1 < n + 1 ; cos (n + l - 2r) x then becomes cos (n f 1 — 2s) x, in which
n + 1 - 2i is always positive.
230 SOLUTIONS OF HENATE-IIUUSE PROBLEMS. [1851.
3. Given f[x) ^f{y) =f{x+y)[\ -f[x)f{y)], find the
form oi f{;x).
Since f[x) +f{y) =f{x + y) [1 -f{x)f{y)} (1),
put X = y = 0^ then
2/(o)=/(0){i-7(om;
therefore either /(O) = 0, or
1 -/(0)> = 2,
giving /(0)=±(-l)i
Taking this latter value, and putting y = 0 in equation (1),
f{x)±{-lY-=f{x){l + {-l)if{x)},
which gives f{x) = + (— 1)^ for all values of x ; therefore the
given equation is satisfied hj f{x) = ± (— 1)*.
Again, if /"(O) = 0, in equation (1) wi'ite y = — x, then
f{x)+f{-x)=f{0){l-f{x)f{-x)],
.■.f{-x)=-f{x).
Differentiating (1) with respect to y, considering x constant,
therefore, putting y = - x^
/(-^)=/(0){i+yW1-^l;
or putting — x = z^
/(^) =/'(o) {i+TRi'l.
Now/'(0) = some constant, C suppose,
.-. ^^ = a{i+./>)l'^
Whence f{z) = tanCs,
or f[x) = tanCic,
which determines the foim of /(ic), C being an arbitrary constant.
1851.] DIFFERENTIAL EQUATIONS. 231
1851.
1. Let P (fig. 98) be any point in a curve and 8 a given
point in a straight line A8\ draw SU perpendicular to SP^ and
let the tangent at P meet SU^ 8 A respectively in U and T\
find the nature of the curve when 8U bears a constant ratio
to 8T.
Let the curve be referred to P as pole and 8A as prime
radius; then
8U: 8T:: sm8TU: Bm8UT
:: 8m{d +8PU): cos 8PU
: : sin ^ + cos ^ tan 8PU : 1 ;
.'. sin^ + cos^ tan 8PU= constant, e suppose.
Now tan 8PU= r -=- ,
dr '
.'. rcosu -— = e — fund,
ar '
cos^ de _ 1
e — m\d dr r '
and log(e — sin &) = log - ,
• a ^
or sma = e — .
r
Transforming this equation to rectangular coordinates,
y = e (a;' + /)*-c,
or eV + (e' - 1) / - 2ci/ - c' = 0 ;
shewing that the curve is an ellipse or hyperbola according as
e is > or < 1.
2. The equation c — a cos^ cos0 — h ainO Hin<f) = 0 may be
considered as the complete integral of
dO #
{d' cos'^ d + J/ sin"^ d - c') ^ ' [d' cos'^ ^ + h' sin'^ 0 - c^
232 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
If c — acos0 COS0 — b siii^ sin(^ = 0 (1),
(a sin 6 cos <p-b cos 6 sin 0) c?^ + (a cos ^ sin 0 — ?> sin 0 cos 0) (^Z0 = 0,
and (1) may be considered as the complete integral of this
equation, or of
acos^ sin^ — Jsin^ COS0 asin^ cos^ - Jcos^ sin^ •••\ n
c being considered an arbitrary constant.
Now (acos^sin^— J8in^cos0)'''+c''' = (acos^sin^— Jsin^cos^)^
+ (acos^cos^ + Jcos^sin<^)'* by (1)
= a* cos^ 0 + b^ sm^(f>j
.'. a cos<9 sin^ - J sin^ cos^ = [a^ cos''^^ + ¥ sm^<f> - c'"')* ;
and similarly,
a sin 6 cos<f> — b cos ^ sin0 = (a''' cos'^^ + b^ sin^^ — c''')*,
therefore equation (2) becomes
[d^ cos^<^ + h^ sin"'^ - c^- [a sin ^ cos ^ — b cos^ sin</>)*
( 233 )
DEFINITE INTEGRALS.
1849.
1. Shew that
^ 7, ax = ir.
+xy
Putting X = tan'' 6, we get
f ^l[££^ dx = 4. r sin'^^ log tan^t^^.
Now generally 1 f[x) dx = I f{a — x) dx^
•) 0 •'0
.-. [ sm'^ log tsinOdd = ( " cos'^ log cot^(/^
•^ 0 •'0
= - 1 cos'^logtan^o?^.
Adding these equal quantities and dividing by 2,
r^log^ db = - 2 [ 'co82^ log tan^ J6>.
Now, integrating by parts,
- /co82^ log tSinOdO = - ^ sin 2^ log tan^ + 6 + G,
.-. - I co82^ log tan^c?^ = i^r,
r" a** logic ,
2. Shew that
I log .r log \^-^) ^-^ = 7r« (l«^g ^ - 1) •
234 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
Let I logic log f — 2 — j dx = w,
dii f lofiTiC
then -^ = 2a \ ., ^ — r, dx:
da J^ x' + a' '
put X = a tan 0, then
dti
^ = 2 T" log (a tan ^) ^7^ (1).
Now generally, I /(a?) <;/.c =1 f{a—x) dx,
■Jo ■^ 0
.♦. ^ = 2 riog(acot6')rZ^ (2).
(1) + (2) gives, dividing by 2,
J /•if
a it
da
= 2 1 log a I
•' 0
= Trloga,
.*. u = 7ra (loga — 1) -I C.
And when a = 0, ?« = 0,
.-. c = o,
and u = ira (loga — 1),
which was to be shewn.
3. Prove that
jg (l + 2ecos04e) ^ ^ -^ ~ 1 - e"
the upper or lower sign being taken according as e is less or
greater than unity.
We have
r 2e + (1 + e") COS0 ^ _ sinO
j (1 + 2e cos 0 + e'f ~ 1 + 2ecos0+e' '
r 2e+ (1+e') COS0, ,^ , ^^ _ , ,. ..
,, . ^^ sS— 2N2 loge (1 + 2e cos^ + e') (?0
J^ (l+2e co80 + e')' *' ^ '
= , — 7i n ^z log 1 + 2e cos 0 + c') + 2e ,- — j^ sr, dB
l + 2ecos0-fe' ^^ ^ /, H-2ecos0 + ey
1850.] DEFINITE INTEGRALS. 235
sin 6/ r cos 9
= ... +
-_ 1 £
1 + 2e cosO + e^ j„ 1 + 2t; cost/ + e'
I —-^ Ti 7. , between the limits,
j„ 1 + 2e COS0 + e" '
W 7fl 1 + «' r"
^0
+ e' + 2e COS0
I + e' r dB
2e Jo 1 + e' + 2e cos0 2e
J0
And
io 1 +
e' + 2e cos^
T. tan~^ tan^TT = ^ if f < 1
1 - <?' 1 + (. ^ 1 _ e
tan"
- tan^TT = g _ if e > 1;
e' - 1 e +
therefore the definite integral becomes in the two cases
I + e' TT IT 2e' , 2
+ — :^ — t •, — TT = TT - — 73 5\ and — TT
- 2e 1-e' 2e 2e(l-e") 2e(l-e')
= + TT
1 - e
1850.
1. Shew that
la '
/ 3
I tan X log (tan a?) dx = Jtt'*',
•' 0
(;8). I -L— — ^ (7a: = Tratans or Trocots,
and that
sm2£
according as s is < or > ^tt.
(a). We have, a being < 1,
,x n-\
(See Gregory's Examples^ p. 477.)
236 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
Diftercntiatiug with respect to a,
/
J o
Z" , , TT COSaTT
Logzaz = —
1 + z ° sm aTT
Putting 0 = tannic, the limits of x will be 0, ^tt, hence
. r*"^ 2<.-i 1 /.. \ 7 tt'^ cos aTT
4 I tan X log (tana;) ax = ^-r, ,
-' 0
whence, puttmg a = §,
27r
TT cos
4 I tan^a; logftancc) £?a: = ^ = ~^ 5
sin-
/
tan X log (tana;) c?a; = ^ tt''*.
(/8). Putting a; = a siu^, we have
^o {a'-x')i ^^ _ ^ p- cos'-'^
— sin^
-I—- — a; " -r-— sint/
sm2s sm2£
therefore wTiting — 6 for 6^
= a —————— rfcr.
- ^— — + sin 0
smzs
Adding these equals and dividing by 2,
J-. « . 8m2ej_w 1 . o^
— « ^ . ..„ sm
sin2£ sm 2£
= _^ r /"i _ cos- 2s
sin2£ i_i^ V (1- sin"''2£sin-'6') '
Tra cos'''2e /'^'^ sec'^^
r" sec
J_w 1+cos'i
« . „ I ^r 5^r— ^ f?^.
8in2e sin2e J i H-cos''^2£tan*^
f sec^ ^
Now j 1 ^ co8''2£ tan'' 6/ ^^ "" ^^^^^ tan"' (cos 2£ tan^) + C;
/•'^ sec'-'^
•*• / ■r~, 27r~z — 27i "" = 7rscc2e;
./_j^l + cos'' 2s tan'^ '
1850.] DEFINITE INTEGRALS. 237
r" {a'-x')i J I 1 co32e\
.*. / ^ ax = irax -r—- ; — —
J _„ a \sm2E 8in2£/
sin2£
= Tra tan s.
The above investigation holds if s < ^tt. If e > ^tt, let
e = ^TT — e' (s' being < ^tt), then
sin 2s = sin2e',
and I ' dx = 7ra tans',
sin 2 s'
= ira cots ;
.'. I 1 ' dx = TTci tans or ira cote,
J-a «
^-— -re
sin 2 s
according as s < or > ^tt.
1851.
1. If 3/ be a function of a: defined by the equation
a^" = {y — nx)"'^^ [y + nxf~^^
shew that f -^ = r -.-^ = -i^ log^^tJlf.
}^y + l3x J^/3y + n'x n + ^ ^ a
Since a'-"' = (^ - na;)"^'' (y + nx)"-^ ;
therefore, taking the logarithmic differential,
/ r,\ dy — ndx , „> ^?/ + ndx
^ ' y — nx ' y -\- nx
W[n + ^)[y^-nx) + [n - ^){y - vx-)] dy
2 2 2
y — « ic
+ n [[n — ^){y — nx) — (w + ^)[y + wa;)} <^]
y — w'^oj
dx _ dy ^
' ' y + ^x /% + Ti'^aj '
•• J y + ^x~ J ^y + n'x
238 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
between proper limits of the variables. Now when a? = 0, y = a,
r ^^ _ r dy
" Ja y + ^x~ Ja ^y + n'x^
d{y +
71X)
^y + ri^x + w^ + n^x '
(w + /8)(^ + waj) '
y + ??a;
the required result.
w + yS
log
2. Detennine the value of the definite integral
"x'-'{\-xY-'dx
[x + ayp '
f
Let
1/
X + a 1 + a '
then when a; = 0, ^ = 0, and when a; = 1, ?/ = 1,
X = ^ ;
1 + a - y^
. 1 _ ^ = OL+^Kizl)
1 + a - y ^
I + a
X -\- a = a
t) =^^'-yy'
1 + a- y^
n-x^
\x +
. ii-a^r'^{i-yr\^i + a-y
" [x + ay ofi 1 + a
Also 7 r- = ,, ^ , ,
{x-VaY (1+a)"'
cfe _ d'y dy
X y 1 + a — ?/ '
— 1 + ^ 7
1851]. DEFINITE INTEGRALS. 239
.-. ^7^^"^l^' da: = ,,/ , y'-' (1 -yf-'dy ;
= 1 r(«lI(/3)
a^(l + a)* r(a + /3) '
the required value.*
* See Gregory's Examples, p. 471.
( 240 )
CALCULUS OF FINITE DIFFERENCES.
1851.
1. PT^ pt^ (fig. 90) are two tangents to a curve drawn at
the exti'emities of any chord PS}) passing through the pole S^
and meeting a given line SA in J", f, respectively ; it is required
to prove that the cui-ve in which the sum of the reciprocals of
ST and St is constant has for its equation
^ = 1 + ecos^ +/(sin^)'^,
where y (sin ^)'^ denotes any rational function of (sin^)'^.
Let the angle SPT = 0, PST = 6, SP=r, then
r sin {0 + (f))
ST^ sin</) '
= sin^ cot^ + cos^,
. ^1 dr _
= sm a — 77, + cos tf :
r da
1 . a^^s a
putting We = -
Similarly, writing O + tt for 6,
1 = - sin(^ + 7r) -^ + Ue^^cos(6'+ tt),
= ^m^-^-7^,^^^cos6';
1.1 . ^ /'du(^,^ dn,
" ST
+ s = ™^ (11"-^)--^^ ("--"»)'
= sin'^-T7^ -^
^ /M0+7r-^'e
de V sin (9
1850.] CALCULUS OF FINITE DIFFERENCES. 241
jNow by the conditions of the problem,
1 1
■^™ + -^ = rt constant, c suppose ;
" dd \ sin^ j sin''<9'
.*. ?<o^.^ — ^(Q = s'm6 {h — c cot^),
h being an arbitrary constant,
= b sin 6 — c cos d,
2e
= cos(^ + a), changing the constants.
But by measuring 6 from a proper point, we shall get a = 0,
so that we may write
2e
^^e+ir -^e = - — cos 6*,
an equation of differences, which, when integrated, gives
Uff = - cos^ + Co,
a
C„ being any function of 6 which does not change its value
when TT + ^ is wa-itten for 0.
Therefore we may put
_l4/(sin^r
^' - a '
and om' equation becomes
_ 1 +ecos^+/(sin^)^
"' a '
1
or, smce Ug = - ^
^= 1 + <?cos^+/(sin^)'.
The following Problem in Geometry of Three Dimensions
(set in 1851) has been omitted.
Determine the surface generated by a tangent to a right
cylinder which moves parallel to the base, and with its point
s
242 SOLUTIONS OF SENATE-HOUSK PROBLEMS. [1851.
of contact lying on a helix : shew also that a hyperbolold of
one sheet may be constnicted touching the cylinder along its
base, and such that the required smface and the h}^erboloid
are developable, the one on the other.
(a) Take the axis of the cylinder as that of z : let a be
its radius, and let the equations to the helix be
z . z
X = a cos - , y = a sin - .
Let I, 77, ^, be the current coordinates of the required
surface, the equations to the generating line touching the cy-
linder at (.r, ?/, z) will be
P cos - + 77 sin - = a, t=^z.
c c
Eliminating z between these equations, we get
t . t
^ cos -4-7? sm - = a
c c
as the equation to the required surface.
(/S) It is obvious that the inclinations of a generating line
of the cylinder to a tangent to the helix and to a generating
line of the h}^erboloid, are respectively constant, and that the
arbitrary parameter of the h}^erboloid may be so detennined
as to make these two angles equal to one another, each equal
the angle t suppose. Let a, a' be two points on the base of the
cylinder, indefinitely near together, a/3, a'/3' the generating lines
of the h}^erboloid passing through them. Also let A, A' be
two points on the helix, such that the elementary arc AA'
equals the elementary arc aa'; and let AB^ A'B' be the gene-
rating lines of the helicoidal surface passing through A^ A'
respectively, and make ayS = a'/3' = AB = A'B'.
We may then shew that BB' = ySyS', whatever he the mag-
nitude of a(3. Therefore, if the circle in which the h}'perboloid
touches the cylinder be laid upon the helix in which the heli-
coidal surface touches it, the element of surface between two
1851.J GEOMETRY OK THFUCE DIMENSIONS. 24;{
consecutive generating lines of the former figure may be super-
imposed on the con-esponding element in the latter figure, and
so for the other elements ; the flexure only taking place round
the consecutive generating lines. Hence the two surfaces are
developable, the one on the other.
We proceed to shew that BB' = /3/3', as above stated.
It Is easy to see that 27rc is the distance between the suc-
c
cessive threads of the helix, and therefore - = cot i : hence
o
the equation to the required h}'perboloid is
■x' + f _ i' _ ,
It is manifest that y8, /3' lie upon the same circular section
of the hyperboloid ; the radius (?•) of this section, whose altitude
call 2,
r = a\\ -{■ —A by the above equation
= a ( 1 +
Also )8yS' subtends at the centre of its section the same angle
that Ota' does at the centre of the base of the cylinder ;
.-. ^^' = - aa
a
[ ay8' sin'^X* ,
Again, B and B' lie on the surface of a cylinder with the
same axis as the proposed, and whose radius (;•')
= [a' + AB'^f.
Also BB' is the same part of the tlu'ead of a helix on this
cylinder that AA is of the given helix ; the helices having
the same distances 27rc between the threads, and therefore
their lengths being
{(27rr')'^ + (27rc)''')4 and {(27ra)'' + (27rcy''}4,
or 27r (a''' + ^^'-^ + c')* and 27r(a'' + c7;
r2
244 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
= ( 1 H a ) -^-^ J *•' c = a cotfc,
therefore BB' = bb'^ aud the surfaces are developable, the one on
the other.*
* We are indebted to Mjf. Cayley for the solution of the second part
of this problem.
( 245 )
STATICS.
1848.
1. A unifonu slender rod passes over the fixed point A
(fig. 91) and under the fixed point 5, and Is kept at rest by
the friction at the points A and B: determine the limiting
positions of equilibriiun.
It is evident that the friction must always act upwards, and
the limiting position of equilibrium will be one in which G
is so near A, that the friction is only just able to support
the resolved part of the weight along the rod.
Let AB = a, AG = x, when G is in the limiting position.
Resolving the forces on the rod along it and perpendicular
to its length,
fiB + fiB' = TFcosa,
B- B'=:Wsma;
and taking moments about Gj
B' {a-\- x) = Bx :
whence the above equations become
Bil =— = TFsina.
V a + xj
Whence, by division.
a + 2x
fi = cota,
and X = 1^ i 0 ^)
which is the least possible value for x: G may be as high
above it as is consistent with the leaning of the rod against B.
24:i) JSULL'TIONS OF SENATK-Iiursi': TKoBLliMS. [1848.
If A were lower tluiu 7?, G miglit be as low as we please,
but at no les.s distance from JJ than the above value of x.
2. Four uniform slender rods, AB, BC, CZ>, DA, (fig. 92),
rigidly connected, form the sides of a quadi'i lateral figm'e, such
that the angle ^ is a right angle, and the points B^ (7, i>,
are equidistant from each other: when the whole is suspended
at the angle vl, detennine the position of equilibrimn.
Let ^, y be the coordinates of the centre of gravity of the
system referred to AB^ AD as coordinate axes; and let
AB = 2ff, AD = 2^», and the angle ABD = a ;
.'. {2a + 2b + 4:{a' + h'y\ x
= 2a.a+2(a''+Z»"04[2a+(a' + Z»')4[cos(120°-a)-cos(120°+a)}]
= 2a" + 2 {d' + hy [2a + («'' 4 Irf- 3* sin a}
= 2a' + 2(«'' + ^»''')*(2« + 3i.^').
Similarly,
{2a + 2Z/ + 4 {pi' + W)"^] y = 2h' + 2 [d' + h'f {2b + 3*a).
Hence, if 0 be the inclination of AB to the vertical,
^l^ b^-\-{a'' + by{2h + SKa]
^ a'+ {a^ + hy{2a + SKb)'
3. A string of given length is attached to the extremities
of the arms of a straight lever without weight, and passes
round a small pulley which supports a weight : find the position
of equilibrium in which the lever is inclined to the vertical, and
prove that the equilibrium is unstable.
The inclination of the lever to the horizon will be deter-
mined in this case in the method to be shewn in the next
problem but one, the point G being now the fulcinim, and P
vertically below G instead of above it.
To determine whether the equilibrium is stable or unstable,
let the lever be turned through a small angle ; then the weight
will assume the lowest position it can, and the normal at this
point to the ellipse mentioned in the above problem will be
vertical.
1848.] STATICS. 247
Hence it is evident that the vertical, through the weight
in its displaced position, will intersect the lever on that side
of the fulcrum which is lowered in the above arbitrary dis-
placement : hence the system will tend further from its position
of rest, and the equilibrium is unstable.
4. Two equal strings, of length ?, are attached to the fixed
points Aj B^ and C, Z>, respectively, which, if joined, would
form a horizontal rectangle ; a sphere, whose diameter equals
AB^ is laid symmetrically upon the strings: find the position
of equilibrium and the tension of either string, supposing
l> AC+^irAB.
Shew also how the problem is to be solved when this condition
is not fulfilled.
The centre of the sphere must lie in the vertical line through
the point of intersection of the diagonals of the parallelogram
ABCD^ and each string must lie wholly in the plane through
its points of support and the centre of the sphere.
Let AB=2a, AC=2b^ and the depth of the centre of the
sphere = z ;
or z = {i{l-TraY-b'}i (1).
Let 2' equal the tension of either string ;
.-. 4:T jj^ 7^ J = weight of the sphere (2) :
from equations (1) and (2) 7" is known.
If I = 2b + TTff, 0 = 0 and T = cc ^ the centre of the sphere
being in the horizontal plane ABCI).
If I <2b + ira^ we must suppose the part of the string not
in contact with the sphere to become rigid, so as to support
the sphere above the horizontal plane ABCD. Each string
must still lie wholly in the plane through its points of support
and the centre of the sphere.
248 SOLUTIONS OF SKNATE-IIOUSE PROBLEMS. [1849.
1849.
1. Two unequal weights, connected by a straight rod with-
out weight, are suspended by a string fastened at the extremities
of the rod, and passing over a fixed point : detemiine the po-
sition of equilibrium.
Let G (fig. 93) be the centre of gravity of W and W\ the
two weights, P the pulley : PG must be vertical, and bisect the
angle WP]V'.
Let TFPTr= 2/, WW'=2a: then PG is the normal of the
ellipse, which has WW for foci and 2l for axis-major ; hence
WP : WG : : T^PT^ : WW,
«^' ^^'^-wTW''^^
W
and IV'P = jjj; :^, 21 ;
W+ W '
hence the sides of the triangle WPW are known, and thence
its angles : therefore Z WPG = I WPW is known, and WGP
= TT — WPG — PWG is known, which is the inclination of
TFTF' to the vertical.
2. A smooth body, in the form of a sphere, is divided Into
hemispheres, and placed with the plane of division vertical upon
a smooth horizontal plane : a string, loaded at its extremities
with two equal weights, hangs upon the sphere, passing over
its highest point, and cutting the plane of division at right
angles : find the least weight which will preserve the equilibrium.
Determine whether the equilibrium is stable or unstable.
Let a = radius of the sphere ;
X = distance of the centre of gravity of the hemisphere
from the plane of division ;
W = weight of the sphere ;
to = weight required.
We may consider the string to become rigidly attached
to the sphere without disturbing the equilibrium : we then have
1849.] STATICS. 249
each system of a hemisphere and weight attached prevented
from turning about the line of intersection of the horizontal
plane and plane of division, by the tension iv, at the highest
point of the hemisphere.
Hence, taking moments about this line,
w.2a = Wx -f ioa\
.-. w = -W=lW (1).
To consider whether the equilibrium is stable or unstable.
If we give either hemisphere a small angular displace-
ment {B) about the above line, the weight to rises through
a space a^, and the centre of gravity of W falls through a space
x.d. Hence the common centre of gravity of the hemisphere
and weight rises through a space
toae - WxO = 0^ by (1),
and the equilibrium is therefore neuter.
3. A slightly elastic string, attached to two points in
the circumference of the base of a right cone, at opposite
extremities of a diameter, is just long enough to reach over
the vertex without stretching. The cone is suspended by it
from its middle point: find approximately the increase of its
length.
Let 2? = mistretched length of the string ;
h = height of the cone ;
a = radius of its base ;
z = the depth through which the cone falls ;
2 (? + \) = the stretched length of the string.
Then, by the principle that " tension varies as extension", if
T be the tension of the string,
T = E J , E & constant weight ;
250 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
h -\- z
and 2 T j — — = W the weight of the cone.
t -J- A,
Also, (h + zY = {l + \y + a';
I I + X
or, omitting V and the higher powers of \,
W f
and A, =
2E {r-dy
4. An equilateral triangle, without weight, has three miequal
particles placed at its angular points; the system is suspended
from a fixed point by three equal strings at right angles to
each other, and fastened to the comers of the triangle : find
the inclmation of the plane of the triangle to the horizon.
Let J, y, 2, be the coordinates of the centre of gravity of
the three weights refen-ed to the strings as axes : ^, y, 2, will
be subject to the condition
X -]ry -\- z = 1^
if I be the length of the strings.
Let 6 be the angle between the nonnal to the plane of the
triangle and the line joining the centre of gravity with the
origin, which is vertical ; this angle will be the required in-
clination of the plane to the horizon. The direction-cosines
of these lines are -,. —,. —,. and j=-^ — z=r. — =:^vi , T=i — 4 — ^svi ?
f^qr^T^^' respectively;
l_ ^ + y + 2
•• ''''^''- ^i' [x' + f + zy
- 1 /
~ '6^' {x'+f+zy
1849.] STATICS. 251
5. A piece of string is fastened at its extremities to two
fixed points : detennine from mechanical considerations the form
which must be assumed by the string in order that the surface
generated by its revokition about the Ihie joining the fixed
points may be the greatest possible.
By Guldinus' property of the centre of gravity, that curve
will by its revokition generate the greatest surface whose centre
of gravity is furthest from the axis, i.e. is lowest, when the
axis is made horizontal and the plane of the curve vertical.
Now we know the centre of gravity will assmue the lowest
possible position when the string is in equilibrium under the
action of gravity : hence the curve required is the common
catenary.
6. It is required to support a smooth heavy body, in the
form of an ellipsoid, in such a manner, that a given radius
in the body shall be vertical, by means of supports at three
points : shew that if /, ?«, «, be the direction-cosines of the
radius, and the equation of the ellipsoid
2 'i V,
X y z
{• - — I — = 1
then the three points in question must be on the curve of
intersection of the ellipsoid with the cone
ll/^ (^. - ^) + ''^-^ (^ - ^.) + ^nxif ( i - 1) = 0.
We will assmne the normals at the three points to meet
hi some point of the vertical radius.*
The equation to the normal at x\, ?/,, 2,, is
^ yjL ?i '
d' h' c'
* This is an ass'omption : for the ellipsoid will be supported if two of
the normals meet in a point not in the vertical radius, provided the resultant
of the corresponding reactions meet the vertical radius in the same pouit
as the third normal docs.
252 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
and this line passes through the points Ir^ mr^ nr ;
//• — .r, mr — y^ nr — z^
^ ^7 ii ^ suppose .
a^ h' e
witli shnilar equations for the coordinates of the other points
of support.
These equations may be written
/>• - ^ . = X,,
mr -fs = y„
nr 7, s = z'.
whence, eliminating r and s by cross-multiplication,
fmz^ ny.\ (nx, Iz\ /7y, mx\
or, di'opping the suffix, we have
as the equation to the cone on which the three points of support
must lie.
1850.
1. A right cone is cut obliquely, and then placed with its
section on a horizontal plane : prove that, when the angle of
the cone is less than sin"^^, there will be two sections for which
the equilibrium is neutral, and for Intennediate sections the
cone will fall over.
Let ABC (fig. 94) be the section of the cone through its
axis, by the plane of the paper, to which the cutting plane
is supposed perpendicular. Let the trace BP of the cutting
plane make an angle 6 with BC: draw PD perpendicular to
BP^ and draw AEF through F, the bisection of BP.
Let 2a be the angle of the cone; then /.ABP=7r — a — 0,
BDP =e + OL, and APD^^O-a.
1850.] STATICS. 253
Also, let AE = n.EF^ then
_AjE _ APs'mAPE _ 2APsm{e-a)
" " EF ~ FF smBFB ~ BP
_ 2 cos(^ + a) slnf^-a)
~ sm2a '
or n sin 2a = sin 2^ — sin 2a;
.-. sin 2^ = («+l) sin 2a.
If ?i = 3, E will be the centre of gravity of the part cut off,
which will therefore stand on its base in neutral equilibrium, and
sin2^ = 4 sin 2a.
Hence, if sin 2a < j^, there will be two values of 6, each acute,
such that the corresponding cutting planes shall give neutral
equilibriiun. For intermediate sections,
sin 2^ > 4 sin 2a,
and therefore ?i > 4 ;
hence the centre of gravity will lie outside the vertical line PD,
and the section will fall over.
2. The three corners of a triangle are kept on a circle
by three lings capable of sliding along the circle, and the
circle is inclined to the horizon at a given angle : find the
positions of equilibiium.
It is evident that, as the triangle is moved about, its centre
of gravity describes a circle about the centre of the circle, the
positions of equilibrium are those in which the centre of gravity
is at the lowest and highest points respectively of this circle.
The corresponding positions of the triangle are easily found.
3. A smooth cylinder is supported in a position of equi-
librium by a string which is wound m times round it, and then
has its extremities attached to two points A and B in the
same horizontal line. The position of equilibrium being that
in which the coils are separate, shew how it is determined.
254 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
and how to find the length of the string in contact with the
cylinder.
Produce the straight lines of the string to meet, as they
must do, in a point : project these produced parts and the
curved part of the string upon the axis of the cylinder; these
projections must be equal ; but the inclination to the axis of
the straight and curved parts of the string is the same ; hence
the produced parts of the string must be equal in length to
the part in contact with the cylinder. Hence the straight
parts of the string occupy the same position as they would
do if the string, instead of supporting the cylinder, ran under
an indefinitely small pulley supporting a weight. This con-
sideration determines the inclination of the straight part of
the strmg to the vertical.
Let 6 = the inclination of the axis of the cylinder to AB',
(fi — the inclination of the string to the axis ;
Q) = the angular distance from the lowest generating
line of the cylinder of the points where the string
leaves the cylinder ;
2a = the distance AB-
2l = the length of the string.
Then, if we project the line AB and the string upon the
axis of the cylinder, we have
a cos 0 = 1 cos cji ( 1) .
Again, if we project AB and the straight parts of the string
produced to meet as above on the plane of either extremity
of the cylinder, the line AB would be projected into a line
of length 2a sin^, and the string into two lines, each touching
the circular end of the cylinder, and of length /sini/r: and
these lines touching the circular end of the cylinder, they make
with each other the angle tt — 2ft). Hence
a sin 9 , .
coseo = ^— ^ — : (2).
/ sni9 ^ '
185L] STATICS. 255
Also the produced parts of the string each equals half the part
m contact with the cylinder = {7mrr + &)?■) cosec<^.
Hence, from the ahove projected triangle,
(mTrr + cor) cosec d>
tan&) = ^^
r
= (mTT+ct)) cosec ^ (3).
From (1), (2), and (3), we may determine 6 and 0; or the
position of the axis of the cylinder and a : whence the lengtli
of the part of the string in contact is knoAyn.
Another condition is, that the centre of the cylinder must
be symmetrically situated with respect to A and B.
1851.
1. A right cylinder upon an elliptic base (the semiaxes of
which are a and h) rests with its axis horizontal between
two smooth planes inclined at right angles to each other : de-
termine the position of equilibrium, (1) when the inclination
of one of the planes is greater than tan"* -r , (2) when the
inclmation of both planes is less than tan *
b
a
Since the locus of intersection of tangents to an ellipse at
right angles to each other is a circle, the locus of the centre
of gravity of the cylinder, as the cylinder is turned about in
a vertical plane, is a circular arc ; and the centre of gravity
is at the extremities of this arc when the axes of the cylinder
are parallel to the planes. Also these extremities are the lowest
points of the arc when the inclination of both the planes is less
than tan~* j- ; but if one of them be greater than tan"* j , one
extremity is the highest point of the arc and the other the
lowest: hence, in this case, the position of equilibrium is that
in which the major axis is parallel to the plane whose inclination
is least ; and in the former case there are two positions of equi-
librium, viz. when each axis of the cylinder is parallel to either
plane.
256 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
2. A^ Bj Cj are three rough points iii a vertical plane ;
P, Q, Bj are the greatest weights which can be severally sup-
ported by a weight PF, when connected with it by strings
passing over A, B^ C, over A^ B^ and over B^ C, respectively:
shew that the coefficient of friction at B = - loe;, -^^r? •
We may consider each of the rough points A^ B^ C, as cy-
linders of indefinitely small radius : hence, by a known theorem
relating to strings passing over rough smfaces, if 6 be the
angle through which the string is bent at any of the points
whose coefficient of friction is ^, and T^^ T^ be the tensions of
the strings on the two sides of the point, if all possible friction
is being exerted, we have
Let fjbj,^ fjb^, ficj be the friction at A^ B^ and C; a, 7, the
inclinations to the horizon of BC and AB respectively: then,
by the question,
p^ £^^(4-7,^gM^(7-;_3Mcli''+').T.]7 (1)^
Q = e^^Afr-^'.sMBfi^+^J.Tr (2),
i? = £'-,(*'-«). £"c'^'+'MF (3);
.-. (2) X (3) - (1) gives
1 ,_ QB
TT
/^B=-lOg3p^^.
( ^57 )
DYNAMICS OF A PARTICLE.
1848.
1. If a and na be the respective distances of a satellite
and of the Siin from a planet, ^ and mp the periodic times of
the satellite and planet, which are supposed to describe circles
round the planet and Sim respectively : shew that the orbit
of the satellite will always be concave towards the Sun, pro-
vided n be greater than m^.
Let the angular velocities of the satellite and planet respec-
tively in their orbits be called eo and mw; then it is plain
that the rectangular coordinates of the satellite referred to the
Sun as origin and axes rightly chosen, are
X — na coswi + a coswwi,
y = na sinw^ + a sinwio)^.
Now, if the path of the satellite pass at any time t from
being concave to convex towards the Sun, we have at that time
M~ ' dt de dt df ~ '
.'. (n smoit -'r m mima>t) (n sinw^-f ni^ sinmcot)
+ {n cosQ>t + m coBmcot) [n coBwt + nf cosnicot) = 0 ;
.•. n^ + m^ + mn[m + 1) cos(7n —\)(i>t = 0.
In order that this equation may not give a possible value
of f, we must have
n^ + m^ > mn (m + 1) ;
.*. T^ — m[m-\-\) n > — ??«"',
or
or w > m
258 SOLUTIONS or BENATE-HOUSE PT^OBLEMS. [1848.
which 19 tlicrcfore tho condition to be fulfilled, in order that
tlie path of the satellite may be always concave towards the
Sun.*
2. A body of given elasticity is projected with a given
velocity, and rebounds n times at a horizontal plane passing
througli the point of projection: detcnninc the direction of
projection, so that the angle between the direction of projection
and the direction of the ball inunediately after the last impact
may be the greatest possible.
Let a, Kj, a^ ••• ^ni ^^ ^^^^ angles of the first projection,
and after the successive impacts;
.'. tana„ = e tana,,_j = e^ tana,,,^ = ... = e"tana,
if e is the modulus of elasticity ;
, . (1 — e") tana
.-. tan (a - aj = ; , J. , :
" 1 + e tan a
we have to determine a, so that this shall be a maximum.
Taking the logarithmic diiFerential of this expression with
respect to tana, we have
1 2e" tana _
tana 1 + e"tan''a ~ '
.*. 1 — e tan' a = 0,
and tana = -r .
3. If a body be projected with a given velocity about a
centre of force which cc . ,. ^ ,., , shew that the axis-minor of
(dist.)^ '
* Since the above condition assigns an inferior limit to the value of m
(n remaining constant), it manifestly precludes the possibility of a motion
of the satellite about the Sun in a direction opposite to that of the planet
i.e. a retrograde motion as seen from the Sun, -which would clearly require
m to be greater than when its path is merely alternately concave and
convex and not looped.
1848.] DYNAMICS OF A PARTICLt:. 259
the orbit described will vary as the perpendicular from the
centre of force upon the direction of projection; and detennine
the locus of the centre of the orbit described.
Let r be the distance, a the angle of projection :
then h^ = SY.IIZ^ the product of the perpendiculars from the
foci on the direction of projection,
= SP sma. HP sma,
= r (2a — r) sin'^a.
And a is constant since the velocity of projection is so ;
.'. b cc sina,
X r sin a,
GC the perpendicular from S upon the direction of pro-
jection.
Also, if p, </) be the polar coordinates of the centre of the
curve referred to the centre of force as pole, and initial radius
vector as prime radius, p = ae^ <f> = angular distance of the
apse, c = distance of projection.
1
=
1
a
1
-
e 1
COS(f>
c
1
—
-7-"
=
1
a
1
-
P
a
C0S</)
1
X
or p^ — cp COS0 + ac — a'^ = 0,
from which equation we see that the locus required is a circle.
4. Two bodies, -4, B, when acted on by gravity, are pro-
jected from two given points in the same vertical line with
the same velocity, and in parallel directions: shew that if A
be higher than B^ a pair of tangents drawn to 5's path from
any point of yl's path, will intercept arcs described by B in
equal times.
S'2
260 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
For if we join the two points of contact, the chord so formed
will be an ordinate to the vertical diameter through the point
in which the tangents meet : let 2y be this double ordinate ;
then, if h be the height of A above J5, and the equation to
B'» path referred to this diameter be
we have y^ = I'h.
Also, if I be latus-rectimi of the parabola, and a the in-
clination of the ordinates of the above diameter to it, the
horizontal distance between the points of contact
= 2y sin a,
= 21'^ sin a. A*,
= 2im,
is constant.
Therefore also the time of passage between the points is
constant.
5. A body is acted on by a force = , ,. ,^ tending to a
(^dist.j
fixed centre S: shew that in general there will be two direc-
tions, ditFerently inclined to AS^ in which the body may be
projected from a given point A^ with a given velocity v, so as
to pass through another given point B.
Prove also that if t, t' be the times of moving from A to B
in the two cases, either t = t' or t + t' = 27r/u. ( ^ —v^] *•
Let the body be projected from A (fig. 95) in a direction
making an angle a with the distance, so as to pass through B :
SA = a, SB = b, lASB = ^.
The equation to the orbit is
d^u fi _
Tn'2 "1" " e 2 ' 2 — ^1
do V a sm a '
or II = .^ (1— e cos(^-7)} (1),
du fM - fa \
la = -2— 2— --2- . e sm c/ — 7 .
1848.J DYNAMICS OF A PARTICLE. 261
Now, when <9 = 0, u = - .
' 'a
du 1
.*• - = ., ... ., (1 — e COS7) : - cota = .; .^ . 2 e 81117.
a vVsin'a^ "^ a vV sura
Eliminating e cos 7 and e sin 7 from equation (1), we have
u = .. ... ■„ ( ., ., . ■■ ) cos^ cota.sin^;
V a sin a \v a sin a a J a
but when ^ = /9, ^ = r ;
.-. 7 = -5^, (1 + cot" a) - \-^, (1 +cot'''a) - -I cosyQ - - cota sinyS,
a quadratic equation, from which the two values of cota can be
detemiined ; which proves that there are in general two diflferent
directions of projection.
Now (Hymers' Ast.^ Art. 326) the time from A to B can
be determined in terms of the focal distances SA^ SB^ the chord
AB^ and the axis-major; and the velocity being given, the
axis-major is independent of the direction of projection : hence
SA^ ABj BS, and the major-axes of the two orbits, are the
same. Therefore the periodic time in the two orbits is the
same; and also the time from A to B.
If ^, t' be the times of describing AB, and the bodies be
projected so as both to describe the angle ASB^ t — t'. But if
one describes the angle ASB^ and the other 27r — ASB^ t + <'
equals the periodic time in the conic section = — - — , where
A equals the semiaxia-major = - — — — Tap") ^^^'
, _ ji^ SP{2A-SP)
"^ ~ SF'' A '
■■■> + >■ = 2.. {^-^)-\*
* For this solution we arc indebted to Mr. Gaskin.
262 SOLUTIONS OF SENATE-HOUSE PllOBLEMS. [1848.
6. Force varvinsr as y^. — -., . shew that, when the hatus-
• ° (dist.) ' '
rectum is given, an angle = 2 tan"' 5^ , measured from the nearer
apse, will be described very nearly In the same time, whether
the body moves ui an elliptic or an hyperbolic orbit, whose
eccentricities are 1 — a and 1 + a respectively, a being small.
We have
dt ~7'
and r = Z (1 + (1 + a) cos^}"\
in the ellipse and hyperbola respectively, where I is the common
latus-rectum : also h^ = fxl is the same in both cases ; hence
§ = X 'a + (!+«) cos^r
= ^-(2cos'i6' + aeos^)-''
~ h 4 V cos'^i^j
r 4 1 /, /, cos^ \ ,
= -y sec ho [1 + 2a — rr-p^ very nearly ;
4 A ^ \ ~ cos' ^6 J -^ -^ '
.-. t = ^ Jisec^e) {1 ± 2a (sec'''i6' - 2)} d tan^^
= ^T /{I + tan'-'i^ ± 2a (tari*|6' - 1)} d tan 1(9.
Hence, if T be the time of describing an angle /3 from the
nearer apse,
T=^^ [tan 1/3 + 1 tan^iyS ± 2a (^ tan^i/3 - tanlyS)}.
Hence the difference of times of describing this arc in the
two cases
2«7^
= f|l (itan^'iyS-tani^),
which vanishes if /3 = 2 tan~'5% and the proposition is true.
DYNAMICS OF A PARTICLE. 263
1849.
1. If the equation for detennining the apsidal distances in
a central orbit contain the factor {u — a)'\ shew that a will be
a root of the equation
(}>{u) - AV = 0,
where ^[u] is tlio central force.
The differential equation of the orbit will be
nnrl
dd
(1).
Multiply by 2 -^ , and integrate ;
(duS" _ C<f>[u).du ,
The general condition for an apse is, that -7^ = 0 ; and there-
fore the equation for detennining the apsidal distances is
If this equation contain the factor [u — ay, let us suppose that
^/^W^-""=/Wl'.(«-a)';
tlien -^={u-a) .f{u),
d'^u du d (du
^^^ de'~ dd'du'Kde
= f{n)[f{u)^f'{u).{a-a)]{u-a)',
hence m = a is a root of the equation
dd' '
that is, a root of the equation
(^[u) - //V = 0.
264 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
2. A body moves from rest at a distance a towards a centre
of force, the force varj^ing inversely as the distance : shew that
the time of describing the space between ySa and ^a will be a
maximum if yS = — j— .
We have here
d'^x _ fx,
~de~~x'
{^\ =2/* log-,
\dtj ° 03
smce a? = a, when -r- = 0 :
' dt '
_ 1 dx
J {F{x) + C] suppose.
Now, let T be the time of describing the space between
/8a and /S''af then
In order that this may be a maximum, we must have
d^ ^'
.-. n^"-'F'{/3"a)-F'{^a) =0.
1
But F'{x) =
therefore the above condition becomes
^^Z i_-o
log^.)
or H*/3"-^- 1 = 0;
1849.] DYNAMICS OF A PARTICLE. 265
.-.-3 = 4:.
the required expression.
3. A particle is attached to the extremity of a fine string,
which is partially wound round a cylinder of diameter c; if
the unwound portion of the string be kept stretched, and the
particle be projected perpendicularly to its length with a ve-
locity F, prove that the string will be wound up after the lapse
P
of the time -^ , where I is the length of string unwound at the
time of projection.
Let r be the length of the string unwound at the time t
after projection, - 0 the arc which has become covered with
string in that time: then, since the only force on the particle,
viz. the tension of the string, is always pei'pendicular to its
instantaneous direction of motion, the velocity of the particle
is uniform:
da ^ ^
.'. r -rr = a, constant :
at
and at the time of projection
r-= V'
dt '
dd j^
dt
f>
Also, r = I — -6;
c ^\d0
^i^)i=^
andi(^-l^) =C-lVt:
and when < = 0, ^ = 0; .-. 6' = ^/';
J -'-6] =P-cVf..
206 SOLUTIONS OK SENATE-HOUSE PROBLEMS. [1850.
Let T bo the time when the string is all wound up ;
.-. t = T, when 1 -^d = 0;
r
, rp
•• Vc'
4. A particle describes an ellipse about a centre of force
in the focus 8 (tig. 96) ; about S as centre a circle is described,
which 'is cut by the radius vector SP in the point Q ; from Q
a line is drawn perpendicular to the direction of the particle's
motion, which meets the major-axis in R: prove that R is
constant in position, and that QR is proportional to the particle's
velocity throughout the motion.
From P draw the normal PG^ and from S the perpendicular
^91^ upon the tangent; also draw GL perpendicular to /S'P, PL
is half the latus-rectum.
Now QR is parallel to PG ;
.-. SR = ^.SQ
^e.SQ,
by the property of the ellipse ; therefore 8R is constant.
Agam, QR = — . 8Q = ^^ . bQ
PL.8Q PL.8Q 1
~ 8PcosP8Y 8Y 8Y'
and velocity cc -qy-]
.'. QR Qc velocity.
Q. E. D.
1850.
1. A heavy particle is fastened by two equal strings of
given length to two points in a horizontal line, and then whirled
round in a vertical plane ; the velocity is such that, if one of
the strings break when the particle is either at its lowest point
or half-way between its highest and lowest points, the particle
1850.] DYNAMICS OF A PAUTICLE. 267
will still continue to describe a circle: find the least distance
between the point to which the strings are fastened that this
may be possible.
Let I be the length of either string, 6 be the inclination
of the strings to the horizon when the distance between the
points is the least possible. Let V be the corresponding ve-
locity of the particle at its lowest point before the string breaks ;
then [V^ — 2gl &\\\d)^ will be its velocity when the strings are
horizontal.
Now, if one of the strings break when the particle is at
its lowest point, it will proceed in a horizontal circle about
the vertical line thi'ough the point of support of the unbroken
string, if the velocity be such as to produce a centrifugal force
just sufficient to keep the string at the same inclination to the
horizon, or, resolving the forces perpendicular to the length of
the string, if
^ ^ .8in0 = q cos ^,
Z cos ^ ^ '
,cos'^ . .
or l^ = gl -T—^ (1).
^ sm^ ^ '
If one of the strings break when the particle is half-way
between its lowest and highest points, it will proceed to de-
scribe a vertical circle about the point of support of the other
string, provided the velocity (F" — 2^'^ sin^)* be great enough
to carry the particle over the highest point of the circle, i.e. if
V""- 2glsm6 > ^gl.
Now, 6 is supposed to have received its greatest possible
value, and therefore, from (1), F its least possible value ; hence
V -2gl%n\d = Sgl,
or al ^^ — 2ql sin 6 = 3gl:
•^ smU
.-. 1 - sin'^ - 2 sin'^^ = 3 sin^,
or sin* ^ + sin ^ = ^,
and 8in^ = - i + (i + ^)*
2
268
SOLUTIONS OF SENATE-HOUSE PROBLEMS.
[1850.
And the least distance between the points of support
= 2?cos^
= (§)Ml + (21)*}W.
2. If P be the perimeter of a closed curve described about
a centre of force, t the time of a revolution, h twice the area
described in a unit of time, and p the radius of curvature
We have
at the time <, prove that P = h I — .
. -'or
p= rvdt
J 0
-/
-/i
dt
't'de
V
dO ,
r -7- .rar
dr
P
rdr
between proper limits.
[r'-fY
Now let (/-/)i=/(^)
•'• / / t!_ 2u between the above limits
= f{0)d0
•' 0
= /(27r)-/(0) = 0;
• P- [ P^P
-I:
■PJ
jyr dr
dr
dp
1850.J DYNAMICS OF A PARTICLE. 269
r —r dr
clr
P
r'de
_ rwdd
~ Jo P
•' (1
p
''dt
3. If any number of bodies be projected from a given point
with the same velocity in one plane, and describe ellipses round
a central force which varies inversely as the square of the dis-
tance; find the law of force tending to the same centre, under
the action of which a body will describe the curve which is
the locus of the centres of the different ellipses.
Let /Lt be the absolute force, V the velocity, and c the dis-
tance of projection. Then, if a be the axis-major,
i = ?-^' is).
a c fj.
Also the equation to the orbit is
1 _ 1 l-ecos(^-a)^
r~ a 1-e' *'
and om' object is to find the relation between e and a ; for if
Pj (f> be the polar coordinates of the centre of the ellipse,
p = ae, <f) = a.
Now, when ^ = 0, r = c;
1
c
1
or -
c
1
a
1
a
1 — e cosa
1 — - cos<&
a ^
1-^
And from (1) a is constant ; hence this equation shews that the
locus required is a circle ; we may put it in the form
^ - p COS0 = d' (- - -] (2).
0 \c aj ^ '
270 SOLUTIONS OF SENATE-HOUSE PUOBLEMS. [iBf)].
Now, by Newton, Sect. ii. Prop. 7, If F be the force in the
circle,
FcrJ-J-
8F' ' FV '
And fi'om equation (2),
SF.SV=a'c{^--\',
\a cj p
and FV = p + a'ci ) - :
\a cJ p'
r cc —,
' '^^^MHJf'
oc £
y + T-
2— r^
C fl
1851.
1. If a body be acted on by a vertical force so as to
describe the common catenaiy, shew that the force and velocity
at any point will vary as the distance of that point from the
directrix.
The equation to the catenary from the directrix, as axis of x,
which we suppose horizontal, is
X X
y = \c [^- + e%
and the force is wholly vertical ;
cFx
df
dx
dt
and -Y = constant = V suppose.
1851.] DYNAMICS OF A I'ARTICLE. 271
. , dy dy dx
Also -r = -# . -r
at dx dt
.". if V = whole velocity,
X X
and V = ^V {eP -\- €~ ')
V
or the velocity at any point varies as the distance of that point
from the directrix.
Again,
d^y _ d (dy\ dx
W ~dx \dt) 'di
ccy;
or the force at any point varies as the distance of tliat point
from the directrix.
2. Force varj^ing inversely as the square of the distance,
a body is projected from a given point in a direction making
an angle of 45° with the distance, and with a velocity = ii times
the velocity in a circle at the same distance : shew that the
direction of the major-axis will be mialtered when the angle
of projection is increased to cot'\l—n^).
272 SOLUTIONS OF SENATE-nOUSE PROBLEMS. [1851.
The general expressions for the elements of the orbit in
terms of the distance (c), the velocity (F), and the angle (^)
of projection, are
1 _2 __r
a c /x '
e cosa = — — — — — 1, a the apsidal angle,
, . V^c sin/3 cos/8
and e sma = ;
u,
.'. cot a = tan/3 — yvo — = — n 5 •
V c smp COSyO
Now, in the present case,
V^ = n^ (velocity in a circle at distance c)
= w' ^ c = — ^ ;
c c
.*. cota = tan/S
= tan/3
w'' sin/3 cosyS
1 + tan'yS ^
w'tanyS '
2
.-. if y8 = 45°, cota = 1
and if /3 = cot"'(l-w'),
1 l + [i-ny_n'~l-{l-ny_ l + l-w''_ 2
l-w"'* w'''"(l-7i^)~ w'^(l-w''') " ^•■' ~ n''
hence the apsidal angle, and therefore the direction of the axis-
major, is the same in the two cases.
3. A body describes a parabola imder the action of two
equal forces, one tending to the focus and varying inversely
as the distance, the other parallel to the axis : find the velocity
at any point and the time of moving between the vertex and
the extremity of the latus-rectimi.
The resultant of the two equal forces will bisect the angle
between them, and therefore be normal to the parabola : hence
1851.] DYNAMICS OF A PARTICLE. 273
the velocity is constant; and if p bo the radius of cui-vaturc
at P,
- = resultant of the two forces
P
= 2 1^ sin SPY
Ox
_2fjL.SY
~ SJP" '
281^
P = -gY" '
.-. v' = ifl,
and V = 2/A*, the required value.
Again, if S be the length of the arc from the vertex to tlie
extremity of the latus-rectum, the time {T) of moving over it
T=2fiiS.
where y'^ = 4fe,
I + X J
ax
J n
W-TI — ??
- + a; + (/a: + x^
= ^Z log i ^ + [Ix + .r'^)i
= ^Z log (3 + 2.2*) + 2* J (between the limits)
= {log(l + 2i) + 2i|7;
r=2/u,*[log(l + 2i)+2i| I
274 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
4. If the product of the velocities at two points P, Q of the
parabolic path of a body acted on by gravity be constant, shew
that the locus of the pole of PQ is a circle having the focus of
the parabola for its centre.
Let a, /9 be the angles between the axis of the parabola and
SP, SQ respectively; then the equations to the tangents at
P, Q referred to S as pole are
1 2
- = y {cos^ + cos(^ — a)},
- = ^ {cos^ + cos(^-/3)|.
r I
Hence, at the pole of PQ,
cos(^ — a) = cos(^ — y8),
and a does not equal yS ; therefore
and at the pole
2 '
1 2 / a + /3 a-yS
^=Tr^-T-+^'^-2-
4 a /8
= -^ cos- cos-.
Now 8P= I sec'^,
4 2 '
^g = ^sec*^f,
and velocity' at P = 2^.;S'P,
Q = 2g.SQ',
.'. SP.SQ = constant (by the question),
or sec'^^a sec'''^/3 = constant;
therefore the value of r at the pole is constant, or the locus of
that point is a circle about the focus as centre.
1851.] DYNAMICS OF A PARTICLE. 275
5. Force varying as the distance, let P, Q (fig. 97) be two
points in the orbit described by a body round a given centre of
force C, and let PT (the tangent at the point P) meet CQ
produced in T; join PQ and draw TA parallel to PQ meeting
CP produced in A ; draw QA meeting PT in Z/, and CU
meeting PQ in F; in CUtake CB a mean proportional between
CV and CU, then the body will pass through the point P, and
the time of movmg from P to B will be half the time of moving
from P to Q.
Let CP=a, CQ = h, then the equation to the ellipse referred
to CP, CQ a,s axes wnll be
+ -3^ + f.= i (1)
Let CA = a\ CT = h' ; the equation to PT will be
X
y
-+rr=l (2).
a
To express the condition in order that this may touch the
ellipse, take (1) — (2)"'' ; therefore
an equation to be satisfied only by the coordinates x = a, y = 0 ;
therefore
1 1
7' ^
or 0 = - .
a
Also, since ^ Z" is parallel to PQ,
a J, c''
which is evidently the condition in order that QA may touch
the ellipse.
Hence QU, PU both touch the ellipse, and by a known pro-
perty of the ellipse, P is a point in the curve if CB' = CU. CV.
t2
276 b(JLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Again, let ^,, </>, 9„ be the angles C'P, CJR^ CQ respectively
make \A\\\ the apsidal distance, then the time in which the
particle will reach P from the apse
= ltan-(|tan^,),
where a and 5 are now the semiaxes.
Let (/<,, /t,), (^.^, l\) be the coordinates of the points P, Q re-
ferred to the axes of the figure, the equations to PZ7, Q U will be
K K
and -^,x + -^,y=\,
and the equation to CZ7 through their point of intersection
and this line passes through the origin ; therefore
\ = - 1,
and the equation to C?7 becomes
a"-
therefore time from PXo R
= -i |ta»"' (j tan<^) -tan-^ (f *^^^i)
_ 1 a tan^ — tan^,
1 + r? tant^, tan©
6 '
1851.] DYNAMICS OF A TAKTICLE. 277
DYNAMICS OF A
rAKTICLE,
K
K - K
a^
«^
"
h - K
ah
h'
?/
/**
/^(/^-/^J
1" + ,
a rt*
' /^.(^\-^J
6*
^1^2 - ^l^a
ah
av;'
^4
">0
1
^A \K
^4.,
ah'
»
1
d' h'
an expression which only changes sign when the suffixes 1 and 2
are interchanged ; it therefore
= -J jtan-^ [^ tan 6',^ - tan"^ [^ tan<^^
= time from B to Q.
Q. E. D.
6. Two bodies A and 5 revolve in the same conic section
roimd the same centre of force in the focus ; shew that if A and
B be at opposite extremities of any focal chord, B will appear
(to a spectator on A) to move with a constant velocity pei'pen-
dicular to SP^ or with a constant velocity perpendicular to the
transverse axis, according as A and B describe the conic section
in the same or opposite directions.
Let PT, QT (fig. 98) be the tangents at the points P and Q
at the extremity of the focal chord BSQ'j draw Sl\ SY' per-
pendiculars from S upon those tangents.
First J suppose the bodies A and B to be moving in the same
direction about S^ then the relative velocity of A and B per-
pendicular to FSQ
= l\smSPV+ V^^mSQY',
278 SOLUTIONS OF SENATE-HOUSE FHUBLEMS. [1851.
it' \\ aiul r, be the velocities of Ji and B
_J_SY h SY'
~ SY' SP'^ SY' SF
~ \SP^ SF
h
= , ,7 : IS constant.
^ lat. rectum
Secondly^ suppose A and B to be moving in opposite direc-
tions about S^ then their relative velocity pei'pendicular to the
transverse axis
= \\co^PTS- V^cosQT'S
- cosSPY- oD> -^^ on^^' ' - cosSQY'
SP sin SPY ' e SF sin SQ, Y' ' e
= - I — cotSFY--^ cotSQY'']
~e [sF
= - {u cot<^ — u' cot0') suppose.
Let ASP =6, then
2
7
2
If = - (1 +e cos^) [I the latus-rectmu),
A ^A. \ du
and. cot m = ^^ :
^ u do'
du 2e .
.•. i( cot 9 = — Jn— T sine',
and writing tt + ^ for ^, and neglecting the change of sign,
since cotd)' = — ^77 and not , -j^ as above, we find
^ u do u do '
?i cot 9 = -y sma,
whence we see that the above expression gives the relative
velocity perpendicular to the transverse axis equal to zero.
1851.] DYNAMICS OF A PARTICLE. 279
7. A body m moves with a uniform velocity v = — — /i*
in a circular tube whose radius is a, and attracts a body m'
within the same tube with a force = , ,. ,„; shew that if m
and m be originally situated in the opposite extremities of
a diameter and m at rest, the two bodies will meet one another
2aa
at the end of the time
b sin a *
Let P, Q (fig. 99) be the positions of m and m at the time t
after the beginning of motion, and let PCA, QCB be their
angular distances from their original positions ^, J5 at the
extremities of the diameter AGB\ let PGA = 6^ PCQ = (f)y
V
also QCB = - t: join PQ ; then, for the motion of m',
a
fl cos|0
(2a)' mi^^<f>'
V
Also, d = TT — (b t ;
a
•*• de " df '
and 2 '-^ ^ = _ iL .^^^ ^ .
dt df 4«* 3in'^<^ dt '
dt J 4a* Vsin*^)^
V* cos'^'a
sin''^<^
+ C
Now, when < = 0, -^ = , d) = 7r:
' dt a
v^ v" cos'^a ,, ^,
••• a' — ^('+^);
.-. C = SGC'OL- 1,
280 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
(It a b1ii^(/)
d<f) V cosa [1 + sijr^^(sec'''a - 1)}*
V cosa {sec" a - (sec'"' a — 1) cos''^^^]* '
, 2a 1 . _, (scc'^a— 1)* , , ^
and f = 7 — r, -Ti sin -^ — cos*© + 6
I' cosa (seca— 1)* sec a ^^
2a
— — : — sin ^ sina cosi6 + (7(=0).
V sin a
And if T be the time that elapses before the collision,
J. = — ; — Sin sma
V sma
2aa
V sm a
8. A straight rod AB (100) slides between two planes
0-4, (9J9, one of which is horizontal and the other vertical :
then, if a body acted on by gravity descends from rest from
the highest point, down the curve which always touches AB^
the time down any arc : the time down the corresponding
chord : : twice the arc : the chord.
Let the length of the rod AB = c, the equation to the
cui've which always touches AB^ referred to Cx^ Cy^ the ho-
rizontal and vertical axes through its highest point, is
(c — xy -f y^ = c\
We will shew that CD is the curve which has the required
property. Let a;', y' be the coordinates of the point P. Then
time down arc CP — ' -^-
•' 0
chord CP = y-^-^^ , 7 the length of the chord.
1^51.] DYNAMICS OF A PARTICLE. 281
Hence the property in the question gives
ds
^Fi "l '
ds
. f dy . 4 tUs ,
• 1 ^ ^ _£__ ^ _ 2 r"'ds
" {gy'Y dy [gyy dy ^^A^^'
. _3_ ^^_2_ £ r^V7.9
" WJf dy' [gyy-y' Idy'^y^
3 , ds ["'ds ^
"~^' w^Ldy^y^
. 3 , ^' 3 ^ ds_
" 2 ^ dy ^ 2dy'"d^'''
or dropping the accents,
ori + 3^' = 0,
y ds '
log.y + 3log^ = logc;
••• r = (-)''
»-(iy=(-;)'
dy \ '
282 SULUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
.-. x-{- C=- {c^-ff,
and the curve passes through the origin ;
.-. 0=-c,
i s s
and [x — cy + ^^ = c'
is its equation, shewing that the curve which has the required
property is the curve generated as CD is.
( 283 )
RIGID DYNAMICS.
1848.
1. A GIVEN Inelastic mass is let fall from a given height
on one scale of a balance, and two inelastic masses arc let fall
from different heights on the other scale, so that the three
impacts take place simultaneously : find the relations between
the masses and heights in order that the balance may remain
permanently at rest.
Let M be the given mass, h the height from which it falls ;
iJ/, M,^ the other two masses, A^, h^ the heights from which
they fall: then the momenta of the three will be
M{2gh)^^ M^[2g\)^^ M,^[2gh^)^ respectively:
in order that equilibrium may not be disturbed, we must have
sum of momenta of iHfj, M^ = momentum of M^
or M^h^i + M^^ = JfA* (1).
Also, in order that the balance may remain permanently at
rest, we must have
M^^M^ = M (2):
(1) and (2) are the required relations,
2. A cannon-ball is fired at a mark at a place whose north
latitude is Z; shew that in consequence of the Earth's rotation
the vertical plane containing the axis of the cannon must be
inclined at an angle of \ht sin/ seconds to the left of the vertical
plane passing through the mark, t being the time of flight in
seconds.
The Earth's motion (oj) of rotation about its axis of figure
may be resolved into two ; one about the vertical line at the
284 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
place In question, and another about an axis tlirough the Eaith's
centre at right angles to tlie former. The latter rotation will
not affect the relative position of the cannon and mark.
The former velocity of rotation is w sinZ, and carries the
mark round the camion from right to left: consequently the
vertical plane through the cannon and mark will in time t
revolve through an angle to sin ?.^, or faWtlo ^^^^^'^ degrees, if
t be the number of seconds in the time of flight, or through
an angle Yf sin/.f, or 15 sin 7.^ seconds: in order, therefore, that
the ball may hit the mark, it must be aimed 15 smLt seconds
to the left of the mark.
3. An imperfectly elastic homogeneous rough sphere is pro-
jected obliquely, without rotation, against a fixed plane ; if t', i'
be the angles of incidence and reflexion. A, the coefficient of
elasticity for direct impact, and p the ratio of the tangential
force of restitution and compression, prove that
2|0 = 5 — 7A, tani' coii.
Let Rj jR, be the normal Impulses up to the time of greatest
compression and during the whole impact respectively :
Fj F^ the same tangential impulses,
F, V the velocities of the centre of the sphere before and
after impact;
.'. R = Vcost.Mj
and V co3^' = "t? — Fcos« = XFcosz (1).
Also tangential velocity before impact = Fsin/; and at the
time of greatest tangential compression the tangential action F
. F
has diminished the velocity F sin* by the quantity -r^, and has
generated an angular velocity w where
Mk'rj = Fa.
1848.] RIGID DYNAMICS. 285
Wc must also express the geometrical condition that the
point in contact with the plane is at rest, or
. . F
Fsmi — ^-p= a-a.
M
„ rr . . F Fd'
Hence K smi — ^r^ = ^rryr, ;
.-. F= ,^\, J/Fsln?'.
af + h'
Also F^ = {\-^p)F;
. . F
.'. F'sint" = Fsini — ~
M
= Fsm;{l-(l+p)^3,}
(2):
Tr . .a — pk
= V sni I —T, — St
ft' + h""
(2) 4- (1) gives
., tan^ a^ — pF
tan^ = — r t. — ^ :
X ft' + Z;' '
or, substituting §ft'' for A;',
., tan I 5 — 2p
.*. 2p = 5 — 7X tan 2*' cot?'.
4. Two given masses are connected by a slightly elastic
string, and projected so as to whirl round: find the time of
a small oscillation in the length of the string. Give a nu-
merical result, supposing the masses to weigh 1 lb., 2 lbs. re-
spectively, and the natural length of the string to be 1 yard,
and supposing that it stretches ^ inch for a tension of 1 lb.
The tension, and therefore the extension, of the string will
evidently depend only upon the relative motion of the masses,
not upon their absolute motions. Now the relative motion of
the masses will not be affected if we apply at each instant to
= the extended length of the string at tune f,
286 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
T
both bodies [M and M') the accelerating force vf, equal to that
acting upon il/, and in the opposite direction : and if, further,
wc apply at the instant after projection to both bodies the same
velocity, viz. a velocity equal to ilf 's velocity of projection in
the opposite direction to it. But by these means J/ is reduced
to rest : let it be taken as the pole of coordinates. Then, if
— = the uncxtcnded length of the string,
1
u
and T = tension at time ^,
we have, by the principle that ' Tension ex Extension',
1 _ I
T=E ''t !i = E "^IsiIL:^ e a constant weight.
1 M "
Now the equation of il/'s motion is
Let u = u^ — a, a will be very small,
and ^-,^„ + a + ^ _ + _ 1 + 3=0;
dd-" « ' " ^ ^ Viif ^ M'j 7/v; V u^
or, omitting d\
which equation shews that a undergoes periodic mequalities,
whose period
E_(l_ l^
l+T^i^ +
1848.] RIGID DYNAMICS. 287
this is the time of the small oscillations in the length of the
string : h must be determined by the circmnstances of projection.
Ex. Let M= 1 lb., M' = 2 lb., - = 1 yard, and E such a
weight that a tension of 1 lb. stretches the string ^ inch, or
iib. = ^Li^^«i^ = j?-
or^= ^^Ibs.;
.'. y = 27r . , ^ ,.. ^ 1 seconds
= ^'^ /^^32,2 x\ 360 3^^ '''"^^''
('
v^ ' 359 * 2,
where v is the velocity of projection, expressed in yards, of M
or M' in their relative orbits,
seconds very nearly.
" 145\i
5. A rough sphere rolls within a hollow cylinder with its
axis vertical, so as to be in contact with the cm-ved smfacc
and the flat bottom: find the reactions and the fiictions, in
terms of the angular velocity with which the sphere goes
round, and explain the indetenninateuess of the problem.
Let (o be the angular velocity of the centre of the sphere
about the axis of the cylinder;
o)., (u_^, co^ the angular velocities of the sphere about that
axis and two other axes of rectangular coordinates ;
B^^ i?^, B^ and i?'^, i?'^, B\ the mutual actions at the
base and other points of contact parallel to the axes
of Xj y, and z respectively ;
a and r + a the radii of the sphere and cylinder.
288 SOLUTIONS OP SENATE-HOUSE PROBLEMS. [1848.
Thcu the equations of motion arc
M"^ = i?^ + n\ - Mg = 0,
MU'^ = - Ra-Ea^mO
(it 1/ e ,
[6 the angle between the radius vector and axis of x,)
JfF^^ = i2a + -B'acos^,
MT^ ^' = Ea sin<9 - E a cos(9.
dt
The geometrical condition to be expressed is, that the two
points of contact must be instantaneously at rest.
Hence,
horizontal motion of the point of contact of the curve surfaces, or
aa>^ — r« = 0 ;
vertical motion of the same, or
ao)^ sin 6 — aco^ cos ^ = 0 ;
and horizontal motion of the other point, or
aco^ cos 6 + aa>^ sin 0 — rco = 0.
Hence,
aco^ = rco cos^,
ao)^ = rco sin^;
.'. a -7- = r cosd.f— r sin^.w",
dco
(where /=-£),
1848.] RIGID DYNAMICS. 289
a —jJi = r {iiT\u.f+ r cosc/.eu ,
dt
It
dto
Also,
X = r cos^,
y = r sin^;
d^'x
.". -j^ = — r sin O.f— r cos6.q)'\
— ^ = r cos 0.f — r sill 6.q)\
Hence the equations of motion become
- Mr sm0.f- Mr cos^.w' = R, + R\ (1),
Mr cose.f- Mr sin^.o)' = E + R\ (2),
0 = i? + ir, - Mg (3),
MFr cos e.f- Mh\ sin O.ui' = - Ra,' - R! p,' sin ^ (4),
MWr sin B.f + MU'r cos 6. ai' = R/i' + R'ci' cos 6 (5) ,
Mk'rf = R'/i' sin 0 - R\d' cos ^ ... (6),
(4) cos^ + (5) sln^ + (6),
2Mk\f=: [R^ + R',) d' sin 6* - [R^ + JS;) a^ cos (9
= -Ma'rf, by (1) and (2);
.••/=o,
and CD is constant.
Hence (4) cos^ + (5) sin^,
0 = i?^ s\n0 - R^ cos^,
and (1) sin^ — (2) cos^,
0 = R\ sin^ - R\ COS0.
Hence there is no action perpendicular to the plane tln-oiigh
the radius vector and the axis of the cylinder. Let R and R'
be the horizontal pressures in that plane on the base and at
U
290 SOLUTIONS OF SBNATE-iroUSE PROIU.EMS. [1848.
the other point of contact. Then (1) cos^ + (2) sin^ and
(5) co8^ — (4) sin^ give
11 + R = - Mrc^' (7),
and R + E = mKvcS' (8):
a'
(3), (7), and (8) are the only equations for determining i?, R\B^,
and B\.
Tlie indetcmiinatcness of the problem arises from the cir-
cmustancc, that there are more pressures on the sphere than
are necessary to produce the motion required. Thus, there are
the two vertical forces B^ and B'^ to support the weight, the
two radial forces B and B' to curve the path of the centre
of the sphere, and the two, B and B\^ to oppose the tendency
of the sphere to rotate about a horizontal axis perpendicular
to the radius-vector. Hence the above equations contain the
sums of couples of these quantities. Considerations of elasticity,
which prevents all such ambiguities in nature, would remove
them from the solution of the problem.
G. A uniform bent lever, whose arms are at right angles
to each other, is capable of being enclosed in the interior of
a smooth spherical surface ; determine the position of equi-
librium.
Find also the time of a small oscillation when the position
of equilibrium is slightly disturbed.
Since the reactions of the sphere all pass through the centre,
it is plain that the resultant force of gravity upon the lever must
also pass through the centre of the sphere ; hence, its centre of
gravity must lie vertically under the centre of the sphere.
Let C (fig. 101) be the angle of the lever ACB, join AB-,
bisect AB, AC, BC, in 0, D, and E, and in ED take the
point a, such that EG : ED :: AC : AC + BC: join OE,
OG, OD: G will be the centre of gravity of the lever, and
OD, OE, will be pei-pendicular to AC, BC: also 0 will be
1848.] RTfJID DYNAMICS. 291
the centre of the sphere. Hence we must have 00 vertical.
But EG: GD'.:CD'.EC::OE:OD; therefore OG bisects
the right angle 0Z>, and^C, 5C, are equally inclined to the
horizon.
^Vhen the lever is slightly disturbed in its own (vertical)
plane from its position of equilibrium, it will manifestly oscillate
as if it were attached to an axis through 0, and the sphere
removed. Hence we have to find the radius (/>■) of gyration
of the straight lines AC^ BC, about an axis through 0, per-
pendicular to the plane of BC^ CA.
Let A^j, k^, be the radii for AC^ BCj respectively : then
k;' = 0D' + IAD"" = EC + ^AB'
= ¥ + !«•■', if ^ C = 2a, BC = 2h.
Similarly,
K = a' + W,
and l>^-J<±JK^ll±^^^±4^±Vl^l(a + hf.
Again, to find 0G{= Z), the distance of the centre of gravity
of the lever from 0. We have
._ j^^, &'mOEG _ a j-,^sinOED
smEOG ^ TTTh smEOG
2-ah
~ a + V
therefore the time of a small oscillation
_ 2^' {a + hf
7. A section of the surtace of a circular right cone (whose
axis is horizontal and vertical angle 60°) is formed by a plane
pei'pendicular to the slant side, so as to contain the vertex ;
U2
292 SOLUTIONS OF si:nati:- HOUSE i'kouuems. [1848.
shew tliat when the surface so cut off makes small oscillations
21a
about the axis, the length (jf the isochronous pendulum = -—
(whether the elliptic base be included in the surface or not),
a being the length of the pei-pcndicular drawn from the vertex
upon the elliptic base.
Let k be the radius of gj^ration of the section about the axis ;
r being the distance of the element hS of the surface from
the axis. Let hS be projected upon a plane pei-pendicular to
the axis, and hS' be the corresponding elementary sm-face ;
.-. hS' = S>Scos30°;
, _ sa^v
= the square of the radius of gyration of the elliptic
projection on a plane perpendicular to the axis.
Similarly, the radius of gyration of the elliptic base equals
the radius of gyration of its projection on the same plane, which
is the same as the projection of the whole section. To find
this radius we must first find the axes of the elliptic base BC
(fig. 102).
Since the vertical angle BAC=SO°^ we have, if AC=aj
and if h equals the semi-axis minor,
(2^)"''= CF.BE=a.2a',
.'. ¥ = K.
We may now also find Oo, the distance of o, the centre
of the base from 0, the point where the axis pierces the base.
We have
Oo^oC-OC
3i 1
1849.] RIGID DYNAMICS. 293
•*.
06' = jW,
and
Oo" = lOd'--
- W,2
or
Od = ia.
Let a', &', be the semi-axes of the ellipse BDj which is the
projection upon the plane BE of the conical surface, as well
as of the elliptic base BCj
.-. 2a' = BD= CF+ \[BE- CF)
= a + \a = la^
and a = |a,
and h' = semi-axis minor of BG
1
Hence the radius'^ of g)T*ation of BD about Oo'
and the length of the simple pendulum of BD about the axis
17 «■'' ^ ,
= 4X6 W"-^'
= T5« + i«
— 2I/T
— TB<^*
Smce the length of the simple pendulum for the conical
surface is the same as for the elliptic base, it is the same for
the conical sm^face alone and taken with the base.
1849.
1. If a miifonn inextenslble string, in the form of any
continuous curve, be subjected to an impulsive tension at its
extremities, the tension at any point will vary directly as the
velocity communicated to that point in the direction of the
radius of absolute curvature, and inversely as the curvature.
294 iSOLUTlONS OF SENATE-HOUSE I'UOBLEMS. [1849.
Let T be the tension at any point, then T -r- ^ "^ ji ^ y >
are the tensions in dircetions of the axes ; and since the tensions
arc impulsive, wc have
difference of tensions at the extremities of any small arc
QC velocity communicated to the arc ;
or -^ al + I —r« as ec ^- (1).
as as at ^ '
Similarly, |ir+r§*cc| (2),
%'^-^'i'^-% (^)-
dx d'^x dy d^y dz d'^z
rp ^ It ~d? 'dt ~dl It Is'
"^ urx\' Td^ /^v *
d'^x dx d^y dy d^z dz
smce -jT 1-+-?^ T^ + TiF -7-=0.
ds ds ds ds ds ds
Now, the direction-cosines of the radius of cui'vature are
d'^x d'^y d'^z
d£^ ds ds^
d''x\' (d'yV fd''zy] i '
Also, if p he the radius of cui'vature,
1 (fd'xy (dW fd'
Ai
p~ Wds'J "^ [ds'J "^ Us'J ^ '
,'. T Gc velocity communicated to [xyz] in the direction of the
radius of absolute curvature, and inversely as the curvatm*e.
2. The nut of a screw rests upon a smooth horizontal plane,
over a hole cut so as to allow a free passage for the screw,
and the screw descends through the nut by its own weight:
detcnnine the motion.
18-19.] KTGID DYNAMICS. 295
At time t let P be the whole action between the screw and
nut perpendicular to the thread of the screw, which makes an
angle a suppose with the horizon. Then
the whole vertical force on the screw = Mg—Fcosa^
moment of the whole horizontal force = Pa sina,
on the nut = — Fa sina.
Hence, if y = depth of any point of the screw below a fixed
plane,
ft), &>', the angular velocities of the screw and nut,
d'^j/ Pcosa
df
7 2 ^Ca
dt
rrid(o'
^ dt ~ M'
The geometrical condition is, that each two cori'esponding
points of the screw and nut in contact have the same motion
perpendicular to the thread;
dy , .
.*. rto) sma 7" cosa = aco sma.
at
Differentiating this equation and substituting from the above.
u
M
Pa sina
M
1
Pa
sina
doi d'^y _ . di
(O
a sma -j cosa -^ = a sma —r- :
dt dt' dt
PiC sin'^'a P cosV Pd^ sin'^a
whence P is constant, and its value known : by substitution of
this value we determine the three required parts of the motion,
which thus appear to be uniformly jWcelerated.
3. The centre of a rough sphere is fixed ; if another sphere
be placed on the top of it and just displaced, determine the
motion of both spheres.
Let 0, o, (fig. 103) be the centres of the spheres at the time t ;
0(7, oc, the two radii which in the beginning of motion were
296 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
vertical, so that C, c, coincided : then, calling the different parts
and angles, as m the figure, we have for the equations of motion,
for the lower sphere,
MU'^=Fa (1).
for the upper one,
M'^ = B smcf> - FcoH<f> (2),
M'^ = B cos<^ + i^sinc^ - M[g (3),
M'k"^^ = Fb (4);-
and for the geometrical condition we must express the circum-
stance that the spheres roll without sliding ;
... a{(})-e) = b{e'-<f>) (5).
Also we have
X = [a + b) sin^,
7/ = [a + b) coscf).
Takmg (1) ^ — (4) «, we have
MFb~-MTa'^^0-
dt' df '
whence we find
., M¥b .
^ M'V'a ^ suppose.
Hence (5) becomes
a{(l>-0) = nbO - b<^y
a ^ + ^ J.
or a = 7 d> ;
and & = n r d>.
a + no
Now the expression for the vis viva gives us
«.(|)V>ri(.+.)'(f)Vr(f)] = 23/-,,(.+.-,);
1849.] RIGID DYNAMIC'S. 297
and substituting the above values of d and 6\ we get
= 2M'g{a-\-h-y)
= 2M'g[a + h) (l-cos(/))
which may be written shortly
dt
\m^ (1 - cos^) = 7>i'''sin"^^<^
dt 1
f/<^ sin^0 '
and mt +(7 = 2 log tan \^.
The constant C may be determined by supposing ^ to have
a very small value a, when i = 0 ; whence
^ , tani(f>
mt = 2 log \-^ ,
^ tanja '
and tan^^ = tan^a £*"",
which determines </», and thence 6 and ^', in terms of t.
4. What must be the angular velocity of a horizontal
cylinder, in order that a heavy string of given length attached
to it may be just wound up ?
Let I be the whole length of the string, x that of the part
hanging doAvn, /u, the mass of a imit of length ; T the tension
of the rope at the point of its contact with the cylinder. Then
for the equation of motion of the cylinder, and that part of
the rope coiled on it,
{Mk^ + fi{l-x)d^}'^ = -Ta (1);
and considering the part of the string hanging down as one
mass, the coordmate of a fixed point of which (the extremity) is Xj
and 1 — X = (10.
298 Solutions of senate-iioiise problems. [1841).
Now (1) — (2) a gives
[Mk^ + /i (/ - a?) a'} -^- fiax-^, =- figax,
(Px fPx
J/A;" + fi[l — x)a\ -T^ + /^a x -j^ = figa x,
d'^x
or {MP + fild^) -jj = f^ai'x.
dx
Multiply by 2 -^ , and integrate,
.-. [Mk' + fild') ('^Y = fjigd'x' + C:
doc
and ;7- = ^) when a; = 0 ; .'. (7=0;
.-•. [Mh' ^ fild') {^\ =figd'x\
Hence, in the beginning of motion, when £c = /,
dd__\^dx_ {H',g)' I
dt~~ li~dt~ [MW + iiM)^ '
which is the required angular velocity.
5. A heavy rod is suspended from a fixed point by two
inextensible strings without weight, the strings and the rod
forming an equilateral triangle ; if either of the strmgs be cut,
dotemiiue the initial tension of the other.
Let the figure (104) represent the position of the beam at
the time t after the string has been cut ; GN being the vertical
line through the point of support. Hence the equations of
motion will be
.lf^^=rsin^ (1),
Jf ^ = Mg - TcosO (2),
Mk^^ = - Ta sin(6'-f <^) (3).
Also the geometiy gives us
X = 2rt sin^ — a sin^,
y = 2a cos^ + a cos(f).
1849.] RIGID DYNAMICS. 299
Hence, differentiating twice, we find
d'x , ^ d'y ^ ^ (ddV . ,„ ,.d"(i)
-aco8(6> + (^)(^'^
T
= g cos^ — -^cos2^,
by (1) and (2).
Hence we find, by substituting the value of -j^ from (3),
-^{co82^+|j8in"''(^+</))}=5rcos^ + 2a(-^) +acos(^+</))f-^) .
Now, in the beginning of motion, — - = 0, -y^ = 0, ^ = SO"",
and 0 = 90°: let 7^, be the initial tension ;
34
and F = ^a^ ;
the required tension.
„ /I 9\ „ 34
or T^'^-^Mg,
6. A man standing in a swing is set In motion : shew that
he can accelerate the motion and increase the arc of oscillation
by crouching and rising In the swing ; and prove that the effect
will be greatest If he crouch when the swing Is at the highest
point, and rise when It Is at the lowest point of its arc of
oscillation.
Since the ropes of the swing are not supposed to slacken
or bend, we may suppose them to become rigid, and rigidly
connected with the swing.
If the man do not crouch and rise, the arc of oscillation will
be unaltered, the effect of gravity being to accelerate the motion
while he Is descending, and to retard It while he ascends.
300 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
Now, if the man rises when the swing is at its lowest point,
the moment of the force of gravity on him about the axis
through the points of support of the swing is diminished, and
the motion less retarded tlian it would have been if he had
retained a mean position ; hence the swing will rise higher than
it otherwise would : if he crouches when at the highest point
of the arc of oscillation, the motion will be more accelerated
while the swing descends than it would have been if he had
remained in a mean position ; hence the velocity at the lowest
point will be increased on account of his having both crouched
and risen ; hence the arc of oscillation will be mcreased by
such a motion of his body.
It is evident that it will be most increased if he rises at
the lowest and crouches at the highest point of the arc of
oscillation.
In addition to the above reasons w^hy the supposed motion
of crouching and rising will increase the arc of oscillation, is
another, viz. that the principle of the conservation of areas
must hold during the sudden motion of rising at the lowest
point. For during that motion both the forces on the man,
viz. gravity and the upward pressure of the swmg, may be
considered as acting in a vertical direction, that is, normally
to his instantaneous direction of motion. The consequence will
be, that his linear velocity will be increased as he rises, and
therefore approaches the horizontal axis through the points of
support of the ropes.
If the swing be supposed to have mass this effect will be
diminished, since his rising will not so much raise the common
centre of gravity of himself and the swing. This diminution
of the effects of the principle of conservation of areas will be
practically caused by a change of the friction between the swing
and his feet, which will for the mstant retard his motion more
than it usually does.
The above reasoning has, of course, no place as applied to
his crouching when at the highest point of the arc of oscillation,
since he is then describing no areas at all about the horizontal
axis through the points of support.
1849.] RIGID DYNAMICS. 301
7. A circular hoop rests upon a smooth horizontal plane
with a particle at its lowest point, and receives a horizontal
velocity of projection V in its own plane : find the value of V
in order that the particle may just rise to the height of the
centre of the hoop.
DcteiTnine the motion when V is greater, and also when it
is less than this value ; and find the time of an oscillation of
the particle in the hoop when V is small.
Let P be the position of the particle in the hoop (fig. 105)
at the time t from beginning of motion. Let AN= ic, NP=y^
be the coordinates of P referred to A^ its position at projection,
as origin, AM = x. The principle of the conservation of the
motion of the centre of gravity in a horizontal direction gives us
M^ + M'^ = M'V (1).
The expression for vis viva is
Also x — X = a sin0,
y =^ a[l — cosO) ;
dx dx rt ^^ /.,\
.-. -: =- = a cosy -r (3),
dt dt dt ^ ^'
dy . add .
-f = a smO -y- (4).
dt dt ^ '
-Y- = a s'mO -T-
dt dt
From (1) and (3),
(31^ M') § = M'[V- a cose '-^;),
{M+M') ~ = M'V+ MacosB^.
ince equation (2) becomes
IM^' H' ( ''- " -^^ I)' + (^' '^+ ^o •=-» f )}
302 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [18-49.
■ja
If the particle just rises to the height of the centre, -^ = 0,
when 0 = 00° and y = a\
- r=(^-^/(2,.)..
AVhen V is greater than this vahie, equation (5) gives the
fid
height to which it will rise before -r- = 0, viz.
y ~ M+ M' ' 2(7 ■
At this time the particle is moving horizontally with the
same velocity as the hoop : it will now fall down in a parabolic
path and will strike the hoop at a point at the same distance
below the horizontal diameter as the point at which it left the
hoop is above it.
If V is not sufficiently great to make it rise to the height
of the centre, it will rise to the height
M' V^
and since the above equations apply for both directions of
motion of the particle, we see that it will continue to oscillate,
rising on both sides of the vertical diameter to the above height.
(19
If V be very small, Q and -j- will be very small : in this
case differentiate (5) ;
-P — ^f^ ] cos''0.2 -^ -r-2 smO cosB -^ i = - 2q -^
\I+M [ dt dt \dfj ) •' dt
' M+M'
or
M'a \, d'^d fddV)
ifTM'f-^^W-^[dt.)\='-^^^
= — 2ga smt; -j- ,
3f +
dO
or omitting OP and 9 ^ ,
1849.] Kin 1 1) DYNAMICS. 303
df ^ M' a'^ '
the equation of oscillator}^ motion, the time of oscillation being
/ M' a\i
8. A heavy lamina, in the form of an equilateral triangle,
suspended fi'om a fixed point by three equal strings, is drawn a
little aside from its horizontal position (the strings being all
stretched) ; its centre of gravity then receives a small honzontal
velocity of projection perpendicular to the plane in which the dis-
placement was made, while at the same time a velocity of rotation
is connnunicated to it in its owti plane : detemiine the motion.
The motion of rotation of the lamina in its own plane is
evidently independent of the motion of its centre of gravity, and
will continue uniform.
By the principle of the superposition of small motions, the
oscillations of the centre of gravity in the two perpendicular
planes will be independent of each other.
The equations of these small motions will be
6 = 6^ cosnt
and (f) = ^^ sinnf^
6^ and ^, being the semi-arcs of oscillation ;
and the centre of gravity will move veiy nearly in a small
ellipse about its position of equilibrium as centre, with axes
rB^ and r(^^, r being the distance of the centre of gravity from
the point of suspension.
9. A man hangs by a rod which swings in a vertical plane :
compare the exertion required to raise him from one given point
of the rod to another, 1st, when he draws himself through a
given small space always when the rod is vertical, and 2ndly,
if he makes the effort when the rod is at its greatest inclination
to the vertical.
See Prob. 6, 1849.
304 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
Let V be the velocity of tlic man at the lowest point before
he begins to raise himself, 2a the arc of oscillation: then, if
a be his original distance from the point of support,
v^ = rt (1 — cosa).
Now, if he raises himself always through the given small
space B at the lowest point of the arc, the velocity at the lowest
point will receive a sudden increase; and it will be with this
increased velocity that he will next airive at the lowest point:
the arc of oscillation will also continually increase.
Let v^ be the velocity with which the man arrives for the
r**» time at the lowest point of the arc, when he raises himself
through the r^^ small space 8. The exertion of doing this
i a-{r-l)8j
To detennine v^. The relation between v^ and v,._^ is given
us by the equation
v^{a-{r-l)S]=v,._^{a-{r-2)B],
which expresses the conservation of areas during the man's rise
through the small space S. We may hence deduce the equation
v^ [a— (r—1) B] = va.
This equation we may also derive from the consideration,
that the man returns each successive time to the point when
he raises himself with the velocity with which he quitted it;
and therefore we may consider that he raises himself by one
effort through the space (r— 1)S, the conservation of areas
holding all the while. This consideration gives us the above
equation immediately. We thus have
a-{r-l)8 {a-{r-l)BY'
Hence the r"^ exertion
Let us call (r— 1)S, x', then we may call S, cfe, and we
shall have, as an approximation to the true result, supposing
B extremely small,
1849.] RIGID DYNAMICS. 305
whole exertion
= M\gh + \vV[j^,^^-l:)
If the man raises himself at the highest point of the arc,
the arc of oscillation will remain mialtered, but the velocity
at the lowest point will be increased after each effort. Hence
every exertion will be the same, viz. Mg sin a. 8, and the whole
exertion Mg sina.^-
Hence the ratio of the two exertions
/ v^[2a-h)\
= I' + ^FF^v '"'''°'-
10. A semicircular board, moving in its own plane without
rotation, and with Its cm'ved boundaiy foremost, comes in con-
tact with a smooth fixed obstacle : determine at what point the
impact should take place in order that the angular velocity
generated may be the greatest possible.
Let P (fig. 106) be the point where the Impact should take
place, the radius CP making an angle 6 with CD the bisecting
radius. Let P^be the velocity (in CD) of the centre of gravity
of the board before impact, t" , v\ those parallel to CD and CB
after Impact, ^ the angular velocity after Impact, R the Im-
pulse : then, if G be the centre of gravity and C6^ = a,
V = V — ■zTi.coaO.
M '
^' = ^«in^> ) (A),
_ Ra %md
306 SOLUTIONS OF SENATE-HOUSE PIIOBLEMS. [1849.
Also we have the geometrical condition, that P must have
no motion in the direction CP after impact ; whence
V co3^ - v sin^ - r^PO amGPC = 0,
or V cos^ — v sin6 — wa sin^ = 0.
Finding r, v', from equations (A) in terms of w, and substi-
tuting in this last equations, we get
fV- — cot^] cos^ - — sin^ - ^'x sln^ = 0,
or V cos^ — ( — ^— ^ + a smO ) •cj = 0
.'. CT =
V sinO COS0
h a sm y
a
ich
I is
to be a
maximum by the variation of 0.
N(
DW,
a
4
= 37r'
a,
k'
= W
-«'^;
.'. in
Va
sin0 COS0
- a' cos'^0
Va sin 2^
«' - a' (1 + COS 20)
Taking the logarithmic differential,
a' sin 20
0 = cot20 -
«'-a'(l + cos20)'
or (a' - a') cot 20 - a'' cosec20 = 0,
or cos20 =
a — a
16
97r' - 16 '
which determines the position of the point P.
11. A imiform solid cylinder is revolving with a given
angular velocity about its centre of gravity, which is fixed ; the
1841).] KKilD DYNAMICS. 307
cylinder then receives a blow of given intensity in a direction
perpendicular to the plane in which the axis moves: determine
the subsequent motion.
Since any section of the cylinder through its axis is a prin-
cipal section, the blow takes place in a principal plane, and
therefore only generates a velocity about the axis (that of
X suppose) perpendicular to the axis of previous rotation (that
of j/), and the axis of the cylinder (that of 2).
Hence, if A be the moment of inertia about the axes of
X and y^ C that about the axis of z ; and w,, eo^, Wg, be the
angular velocities about these respective axes at any time t
after impact, we have as equations of motion,
A'^-{C-A)a>.^^o.^ = 0,
C^-{A-A)co^<o^ = 0,
or ^^ = 0,
df '
and 0)3 = constant
= 0, since that is its original value.
Hence also tUj = constant
= that generated by the blow
= — 7- , if i? be the blow, a the distance from
A ' '
the centre of its point of application,
and (o,, = constant
= its original value before impact.
Hence the cylinder revolves unifonnly, and the instantaneous
axis is fixed in it, viz. in the plane xy ; hence this axis is also
fixed in space, and the axis of the cylinder, as before, sweeps
out a plane.
x2
308 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
12. A Rolld cone is suspended by its vertex from a point
in a perfectly rough wall : if the cone be slightly displaced from
its equilibrium position, the surface remaining in contact with
the wall, determine the time of a small oscillation.
Let 0 be the angle which the line of contact of tlie wall
and cone make with the vertical at the time t] to the angular
velocity at the same time about the line of contact ; 0^ the
original value of ^; h the height and 2a the vertical angle
of the cone : then, since the line of contact is the instantaneous
axis, the equation of vis viva gives us
Jf/c'V = 2Mg.^h cos a (cos0 - cos^J,
2 3 qh cosa , ^ n\
or <" = n 7,>ii — (cost/— cos^J.
Now, to connect w and 0, we have two expressions for the
rlB
motion of the centre of the base, viz. h sina.w and h cosa . -r- ;
dO ^
.-. 0) = ^ cota;
fd()\^ 3 qh sin^a , ^ n\
•■• [it) "2 'ks^f'^''^^ -"><);
.*. -jY = - J \n sin0 , I the length of the side ;
or if the motion be very small,
d''0 3 «7/sin'''a „ ^
k'
Therefore the time of a small oscillation = 47r
(3^Z)^ sina
13. An indefinitely great number of indefinitely thin cylin-
drical shells, just fitting one within another, are revolving with
different angular velocities, but in the same sense, about their
common axis; also the angular velocity of each shell is pro-
portional to a positive power (the n"^) of its radius, and that
of the outermost shell is tu. Prove that if the system of shells
1849.] RIGID DYNAMICS. 309
be suddenly united into a solid cylinder, the cylinder will revolve
about its axis with the angular velocity .
If the bodies composing the Solar system were suddenly to
become rigidly connected, explain what the nature of the sub-
sequent motion would be.
If 0)^ be the angular velocity about the axis of any particle, at
a distance r from the axis, the area described by it in the time t
n n+!j
7' r
a a
if a be the radius of the outer shell.
Hence the sum of the areas described by all the particles in
the same cylindrical shell, of thickness Sr,
= TTO) -%- t . Br.
a
Hence the sum of all areas described by all particles in time t
t r
— '
« Jo
a
= rrmt
« + 4
If w be the angular velocity after uniting, the sum of the
areas described by all the areas in an equal time
= TT'ot I r^hr
•' 0
= TTWC . —
4
By the principle of the conservation of areas, the two above
sums of areas must be equal, or
a' or
TT-art — = TTtOt 1
4 n + 4 '
4(i)
« + 4
310 soLLTioNS OF sENATK-iiuL'sE rKOBLi::>:s. [1850.
In the Solar system the areas of the orbits vary as the square
of the mean distances, and the periodic times squared as the mean
distances cubed, and therefore the mean angular velocities vary
as the square roots of the mean distances. Hence, if all the
bodies of the system were suddenly to become rigidly connected,
the bodies nearer the sun would have their motions suddenly
accelerated, and those furthest from it would be suddenly re-
tarded ; after which all would proceed with a common unifonn
and, so to speak, an average angular motion.
1850.
1. A parallelogram, whose centre is fixed, is rotating about
one of its principal axes in its plane ; find how it must be stiiick
that, after the blow, it may rotate with the same angular ve-
locity about the other.
Since the effect of a blow upon the velocity of rotation of
the body about a principal- axis depends only upon the moment
of the blow about that axis, it is plain that the blow must in
this case be perpendicular to the plane of the parallelogram,
and Its moments about the two principal axes in its plane must
be equal and be due to the velocity of rotation already existing
about one of them, and in a direction to destroy it.
Let A, B^ be the moments of inertia about these principal
axes, 0) the velocity of rotation about A before the blow (/)
IS given, x, «/, the coordinates of the point of application of the
impulse ;
Am
y
Bm
which equations determine the point of application of the im-
pulse.
2. Three equal smooth spheres (radius r) are placed together
on a horizontal plane, and kept in contact by a string passed
round them in the plane of their centres. A cone of given
weight [W] and vertical angle (2a), is placed between them
1850.] nroiD dynamics. 311
80 that its axis is vertical : find the tension of the string ; and
if the string be suddenly cut, find when the cone will strike
the plane.
If R be the pressure between any sphere and the cone, we
have for the equililibrimn of the cone
3i2sma = TF;
and for that of any sphere,
i?cosa = 2 TcosItt
„ FTcota
the required tension.
At the time t after the string is cut, let y be the height of
the vertex of the cone above the plane, x the distance of any
sphere fi'om the axis of the cone : the equation of vis viva gives
us, if W be the weight of any sphere,
where y^ equals the height of the vertex of the cone in the
position of equilibrium.
We have also to express the condition, that the motion of
the point of any sphere in contact with the cone in the direc-
tion perpendicular to the generating line of the cone through
the point of its contact with that sphere, is equal to the motion
of that point in the same direction : or
dx dii .
T- cosa = — J- sma :
dt dt '
dx dy ^
.'. -J- = r tana.
dt dt
Hence the above equation becomes
( W+ 3 W tan'^a) (^)' = 2 Wg {y, - y) :
or differentiating,
^ Wg_
de ~ Tr+3ir'tan''a
= — /' suppose,
312 SOLUTIONS OF SENATE-HOUSE FKOliLEMS. [1850.
a constant retarding force. And the cone starts from rest;
therefore the time of describing the space y^,,
To find y^. From fig. 107 it is evident that OiV, the per-
pendicular on the axis of the cone from 0, the centre of any
sphere in the position of equilibrium, is the distance of any
angular point of any equilateral triangle of side It from its
centre ;
/--» AT TT 2
.-. 6>A = r sec - = p r ;
.*. OT = ON seca = ^ ?' seca,
and PT = ( -5 seca - 1 J r ;
CT = PT coseca
2
-J seca coseca — coseca j r ;
.-. CN=^ CT-NT=CT- OTsma
I r^ (seca coseca - tana) — coseca^ r
= f-i cot a — coseca] r,
and y^ = r — CN
= f 1 -f coseea — —^ cota j r-;
1(1+ coseca - -j cota) ( W-\- 3 W tan'a) r j
3. Two equal particles of mass m are fixed at the ex-
tremities of the axis of a prolate spheroid, of which the mass
is il/, the eccentricity of the generating ellipse being e. The
spheroid is struck by a couple and then left to move freely;
1850.] RIGID DYNAMICS. 313
shew that throughout the motion it will constantly have contact
with a single plane, if
m = ■^o^e^
The spheroid will manifestly have contact with a fixed plane
parallel to the invariable plane, if it be similar to the momenta!
spheroid of the system consisting of the spheroid with the two
masses [m) at its poles. Let -4, jB, be the moments of inertia
of this system about its axis of figure and an axis through Ita
centre perpendicular to its axis of figure. Hence, if a, 5, be the
semiaxes of the generating ellipse, the condition of contact with
a single plane is
Ad'=^Bh' (1).
Now A = lMb\ B==\M{a^ + h')-\-2ma\
and condition (1) becomes
or 2ma' = lM{a'-b')-
= me\
4. A small arc of a hoop is removed and replaced by two
small straight lines, tangents to the circle at the ends of the arc,
their mass being so disposed that the centre of gravity remains
still at the centre of the hoop. If the hoop be now rolled along
a horizontal plane, sufficiently rough to prevent sliding, with an
angular velocity a> not great enough to make it leap, shew that
motion will never cease unless
. f o)''* d^ + F cos'^a
\2ga ' (1 — cosg) cos a
be a whole number ; where a is the radius of the hoop, k its
radius of gyration, and 2a the angle subtended by the arc
removed.
314 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
The motion of rolling on the circular rim will be unifoiTn :
hence the angular velocity at the time of impinging on the apex
of the tangent will be to. To determine the motion immediately
after this impact. See fig. 1 08.
Let F^ F' be the impulsive actions between the apex and
the plane along the plane and pei'pendicular to it; v^, v^ the
velocities of the centre of gravity in the same directions, and
■C7 the angular velocity after impact. The equations for finding
y^, V . and -nr, are
F
F'
MFzj = Mk^m + Fa - F'a tana.
Also, to express the condition that the apex must be at rest
after impact, we have
v^ — aw = 0,
Vy — a tana.-cT = 0.
Hence we have
Jc^z: = Jc^oi + a^ (o) - w) — (a tana)^.^,
a'(l+tan''a) + k'
= nco suppose.
;Next, to consider the continuous motion of turning about
the apex : let 6 be the inclination to the horizon of the radius
to the apex at the interval t after impact. Then, taking the
equation of moments about the apex,
M{/c^ + c^ sec'^a) -yy = — Mga sec a cos^ ;
'd6\
(da\ 2qa sec a . ^ ^
\dt J F + a" sec^a
2qa seca , . ^ >
7 2, 2 2- sm e - cos a
k -f a sec a ^ '
1850.] RIGID DYNAMICS. 31,')
\dtj k~ + a' sec'a ^ i \ n
a 1 A'^^^'
In passing from a motion of rotation about the axis to a
motion of rolling on the circular rim, there is no impulsive
motion ; hence the angular motion at the time of the second
impact of the apex on the plane is w or noa. Similarly, the
angular velocity at the time of the >•'*' impact is ifca.
Now, if the hoop ever comes to rest it must be by just
balancing on the apex : suppose this happens when it is rolling
over the apex for the ?«"' time; then equation (1) shews that
we must have
= w ft) - ^-^ — ^ — 1 -cosa),
/c + a^ sec'a ^ '^
o 1 if 2(7a sec a ,,
or 2m log?i = log ijj^-^, r^-^z l-cosa
'^ [[k + a sec a) w
ft)" («^ -f P cos''^a)
lof
^ I2qa (1 - cosa) cosa
or m = ^\ - 1
2log|n-^^^tan^a|
by inverting both the quantities under the logarithmic sign.
Hence, if this expression for m be a whole number, the hoop
will come to rest as it is rolling over the apex for the ?»"• time :
if it be not a whole number it will never come to rest.
5. Two similar homogeneous cords are similarly stretched,
and one of them loaded at its middle point with a small
weight /x; shew that the fundamental note of the loaded cord
will be lower than that of the other, and that if t denote the
time of vibration of the loaded cord for any possible note, the
values of t are given by the equation
tan H fiVl = A Uxl<
wlicrc / is the length of the cord, • the mass of a unit of length.
316 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
and T its tension. Under what initial circumstances will both
strings sound the same note?
See Duhamel in the Journ. de VEcole Polytech. tom. xvii.
It is evident that when the loaded cord is sounding its fun-
damental note, its two halves meet in a salient angle pointing
from the line joining its extremities. Thus the two halves are
portions of the trochoid which a longer cord would assume in vi-
brating, and will vibrate in the same maimer as if they actually
were parts of such a trochoid : hence the loaded cord is virtually
longer than the unloaded one, and capable of a deeper note.
Let ic, 3/, be the coordinates of any particle of the string at
the time f, referred to the line joining the extremities of the
string, and a line through its middle point perpendicular to it
and in the plane of vibration, as axes of x and y. The equation
of transversal vibrations will be (Poisson, Mecanique^ n°. 490)
, . rmr ,. , > mir
y = h sm -^— [\L — x] cos ^— at^
A A<
where h is the greatest value of y for any particle, a = ( - ) , and
\ a quantity to be determined by the circumstance that the
middle point of the string is attached to the weight [i.
Let y be the ordinate of this weight at the same time t :
the equation of motion of ^l is
Now
tan
d'y
=
KIL
=
hrriTT rmr I rmr
— 2t . — — cos — - - cos — - at]
A, A 2 A
••• y
=
2t h\ 1 rmr I rmr
— . — . —:. cos — — - cos -T— at.
fi rmr a A 2 A
y
=
Jt=o
=
J . rmr I rmr
h sm — - - cos — - at :
A <S A
rmr I
2t \ 1
\ 2
fj> ' mir ' d^
=
2c \ . , T
— . — , since a = - .
fx. ymr £
1850.J RIGID DYNAMICS. 317
Now, let t represent the time of vibration ;
_ 2X, 1 _ 2\ /£
m ' a m ' \T
and the above equation becomes, by eliminating \,
the required equation for the determmation of all possible values
of «.
The value of t answering to the fundamental note is its
greatest value; it is plainly such that
T-l;) "2-
Now suppose fi indefinitely small, or the weight removed,
the value t' of t then answering to the fundamental note is
TT? /S\4 _ TT
T-[t) -2'
and therefore t > t\ or the fundamental note is lower for the
loaded than for the unloaded cord, as shewn above.
The two cords will evidently sound the same note when the
middle point of each is made a node ; in which case the note
will be that due to a length which is any submultiple of \l.
6. If a body hang by a string, and through any point of the
string a series of horizontal lines be di-awn, with any one of
which the body may be rigidly connected and perform small
oscillations about it, the time of oscillation will be a maximum
about a line at right angles to the one about which it is a
minimum : prove this, and shew how to find the position of these
two lines, and the time of oscillation about any other, in tenns
of the times about these two and the angle which it makes with
them.
If t be the time of the body's oscillation about any one of these
lines about which its moment of inertia is Q^ we have the relation
« = ^^ (j4)''
318 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
where M is the body's mass, h the depth of its centre of gravity
below the horizontal line in question. Now the general ex-
pression for Q is
Q = ^hn{x' + ^f + z');
therefore if we make the horizontal plane through the lines the
plane of xy^ 'S.hmz^ is the same for all the lines, and the relative
magnitudes of Q for the different lines will depend upon
But this expression will be unaltered if we project every particle
of the body upon the plane of xy. Hence, as in the case of a
plane lamina, we shall have two axes in the plane at right angles
to each other, for one of Avlilch Q will be a maximum and for
the other a minimum : hence the first part of the proposition
is tnie.
Also, as far as the part ^hn (.c^ + y"^) of Q is concerned, we
shall have the usual relation
Q = Q, co^'9 + Q,^ sin'6',
where Q^^ Q^ are the maximum and minimum moments, and 6
the angle which the axis of Q makes with that of Q^ : hence the
true relation between Q^ Q^^ and Q^ is
Q - 28m.z' = {Q^- ^Bmz^) cos'0 + {Q,- ^Smz') sm-*^,
or Q = Q^ cos'^^ + Q^ sin^^, as before.
Hence, if #, t^^ t^^ be the times about the lines about which
Qi Qit Qii ^1'® the moments of inertia,
t = {t'^ cod'd + f.; sln'^)i
To find the positions of the lines of greatest and least
moments in the given horizontal plane.
Let the direction-cosines of this plane referred to the prin-
cipal axes thi'ough the point where the string pierces the plane
be ?, m^ n] and let Q be the moment about the line whose
direction-cosines referred to the same axes are a, /3, 7 ; then we
are to have, if ^, B^ C, be the principal moments,
Q = Ad^ + B^^ + C<f = a maximum or minimum
1851.] RIGID DYNAMICS. 319
subject to the conditions
a' + 13' + 7'^ = 1,
and /a + w/3 + ny = 0,
which last is the equation to the horizontal plane.
Differentiating these three equations with respect to a, /9,
and 7, we find
Aada + B^d^ + C^^dy = 0 (1),
ada + ^dl3 + rydy =0 (2),
and Ida + md^ + ndy =0 (3).
Using the arbitrary multipliers \ and //., we deduce the equations
Aa + \a = fil (4),
Bl3 -\-\^ = fZ7n (5),
Cy + \y = fjin (6).
(4) a + (5) /9 + (6) 7 gives
Q + \ = 0',
.-. {A-Q)a =fil 1
{B-Q)^ = f,m\ (A).
{C-Q)y=jjLn}
Hence a^+ ^^+y^=l= /."^ j^-^^ + ^^. + (-^|-(7),
and
1 r m^ w'
_(?« + ,„^ + ,,^) = 0 = ^—^ + ^3^ + ^^,
which gives a quadratic for the detenmination of Q^ and Q^.
The substitution of the value of yu. from equation (7) in equations
(A) will give us the values of a, /3, 7, and so dctcnnine the
positions of the lines of maximum and minimvun moments.
1851.
1. The locus of an axis passing through a fixed point of a
solid body, and such that the moment of inertia round it of the
body is constant, is a cone of the second order, and the cones
320 SOLUTIONS OF RENATE-HOUSE PROBLEMS. [1851.
coiTesponding to different values of the constant moment have
the same directions of circular sections.
The expression for Qj the moment of inertia about an axis
whose direction-cosines referred to the pi*incipal axes are a, /3, 7,
in terms of the principal moments A^ B, C, is
or if the axes of A, B, (7, are those of .r, ?/, and 2:,
Q [x' +f + z') = Ax' + By"" + Cz\
the equation to a cone of the second order.
Let Aj B, Qy (7, be in descending order of magnitude, the
above equation may be put in the form
{A-Q)x' + {B-Q)f - {Q-C)z' = 0;
and if 2 = mx + 7iy + c
be the equation to any plane which cuts the cone in a cu'cle,
we have {Oregory''s Solid Geometry^ Art. 124)
n = 0,
_ ^ \A-Q-[B-Q)\^
'A-B\i
= + '
.B-CJ'
which shews the direction of the circular sections to be inde-
pendent of Q.
2. Determine the motion of a heavy solid composed of two
equal right cones placed together base to base, and which rolls
without sliding upon two intersecting lines inclined at equal
angles to the vertical, the common base of the cones moving
in the plane which bisects the angles between the vertical planes
through the lines.
We shall apply the principle of vis viva.
Let CA^ CB (fig. 109) be the two lines ; and at the time t
let them touch the two cones in P, P' : let GM the height of G
1851.] RIGID DYNAMICS. 321
the centre of gravity of the soHd above tlie horizontal plane
through (7=2;, CM = x. Then if r be the distance of P from
the axis of the cone, and S the inclination of the plane A CB
to the horizon, FN the height oi P = z — r cosS, or if CP = p,
psiny = z — rcosS;
.■. 2; = p sin7 + rcosS (1).
Also, if z MCN = £,
X = CiV coss — pcoay coss (2).
Now as the cone rolls along, the locus of P on the cone will
be a cui've like the dotted curve in the figure : let Bs be an
element of the length of this curve answering to the rotation
of the solid through a small angle 89^ then
Ss — S (rcoseca) cosecyS (3),
if /3 be the inclination of either rod to the common base of the
cones, or 2/S the inclination of the rods to each other : also
8 (rcoseca) = 8[r0) tan/9 (4) ;
.*. 8p = - 8s = — Sr coseca cosec/S by (3) ;
.•. p = (r^ — r) coseca cosec/S,
if r^ is the radius of the common base of the cones : also by (4),
r — r^ = rStan^ sin a,
if 0 = 0 when the solid touches both the lines at (7;
.•. 0 = ( 1 -] cotS coseco-
V rj
Hence, by (1) and (2),
z = (>;| — r) coseca cosec/S sm 7 + ?-cosS (5),
X = (r^ — r) coseca cosec/S C0S7 coss.
Now the equation of vis viva gives us
322 .SOLUTIONS OF SliNATE-IIOlSE PKOBLEMS. [1851.
where «„ is the height of p where tlic solid starts from rest ;
.', {cosec^rt cosec'''/9 cos'^y cos'^'s + (coseca cosec/9 sin 7 — cosS)"''
Pi-:' ,._ , (dry
l'< v^o 1 (dry
+ —^ cotp coseca} I ^-1
= 2/y [z^^ — 7\^ coseca cosec/3 sin7
+ r (coseca cosec)8 sin7 — cos8)},
an equation which, when integrated, will give lis r, and there-
fore also X and 2, at any time t.
From equation (5) it appears that as z will necessarily be
diminished by the force of gravity, if the solid starts from rest
it will roll toward C or from it, according as
coseca cosec/3 sin7 > or < cos 8.
3. Shew that the diiference of the moments of inertia of
a body round two axes in a given plane which are equally
inclined to a fixed line in the same plane, is proportional to the
sine of the angle between those axes.
Let ^j, Q^ be the maximum and minimum moments about
lines in that plane, Q^ Q the moments about any two lines in
the plane making angle 0, 0' with the axis of the moment Q^ ;
then, by Problem 6, 1850,
<2 = ()^cos'0 + <?^sin'0,
.-. Q- Q = Q^ {co&'e-QOs'B') + Q,^ (sin"-"0-sin'^0')
= Q^ sin {B'-B) sin {B + B') + Q,^ sin {B - B') sin (0 + B')
= {Q,-Q,)sm{B-\-B') sin {B'-B)
= ((?,-^Jsin2asin(0'-0),
where a is the angle the fixed line makes with the axis of Q^
oc sin(0'-0),
Gc sine of the angle between the axes of Q and Q'.
{ 323
HYDROSTATICS.
1849.
1. A plane body, one of the edge3 of Avhich is a straight
line, is immersed in water so as to have this straight line coin-
cident with the surface; shew how the depth of the centre of
pressure may be deduced from observation of the time of a small
oscillation in vacuum of the body about its rectilinear side.
WTien the body is immersed with its plane vertical,
let z be the depth of the centre of pressure below the surface,
••• z gravity ,
... z any point of the body
Jgpz'dz
Then z' =
Jf/pzdz
Mz z '
(where h' is the radius of gyration about the straight edge)
the length of the simple pendulum when the body makes small
oscillations about the rectilinear side. Hence, if T be the time
of small oscillations.
r=27r
.'.Z^l^C,—^,
the formula for the detcnnination of z from T.
2. A cylinder the radius of which is a, having its axis
vertical and containing incompressible fluid (density /a), re-
volves about its axis with an angular velocity o) = [»-r,
Y2
324 SOLUTIONS OF SENATE-HOUSE PR(lBLEMS. [1849.
n being > 1 ; a sphere (density p'), whose radius is also «, on
being put into the cylinder, is supported in such a position that
it touches the free surface at its vertex : shew that
p \ n
The equation of equilibrium is
volume of fluid displaced = — .^Tra" (1).
r
Fig. 110 shews the position of the sphere in the fluid, the
dotted line representing the continuation of the section by the
plane of the paper of the free surface.
Let F, the vertex of the free surface, be the origin of
coordinates, the axis of the cylinder that of 2, and r the distance
of any point from it. Then the equations to the free surface
and that of the sphere are
2 2a
CO ^ n ^
and r'"' = 2az — s'"* :
therefore if z' be the height above V of the circle of intersection,
or z' = 2(7 f 1
Hence the volume of the fluid displaced
= irj^{2az-z')dz--.-
TT -I « ( 1 — -] z"^ — !«'■
3'
= W(l-l){a(l-i)-Sa(.-l
1849.]
therefore from f 1 '
IIYDUOHTATrcS.
^.|W = ..a^(l-^,,
325
or
- = ('--)'
> V nj
3. If X, F, Z^ be the forces acting at a point [xyz) of a mass
of heterogeneous fluid in equilibrium, and Xdx + Ydy + Zdz^
be not a perfect differential, then the pressure and density will
be constant thi'oughout the curves of which the differential
equations are
dx dy dz
d]^_d^~ dZ _dX~ dX _dY'
dz dy dx dz dy dx
Let p^ p be the pressure and density at the point (xyz) ;
]) + dp the pressiu'e at the point [x + dx^ y + dy^ z + dz)^ then
dp = p {Xdx ^-Ydy + Zdz) (1).
In order that this equation may hold, and therefore equi-
librium be possible, we must have the right-hand side of this
equation a perfect derivative of tkree independent variables,
or we must have
dp Y dpZ
dz
dpZ
dx
dpX
dy
dpX
dpY
dx '
or
ldY_ _dZ\ ^ ^dp _ ydp^
'^ \dz dy) dy dz
dZ dX\ __ Y ^P ydp
dx dz J ^ dz dx
(A).
dX
('
\ dy dx
dY\ _ Y^ _ x^^
dx
dy.
326 SOLUTIONS OF SKNATE-HOUSE PliOHLEMS. [1849.
^lultiplying these equations in order by A', Y^ Z, and adding,
we get
as the condition wliicli the forces X, F, Z^ must satisfy in order
that they may be able to produce equilibrium.
Now let dx^ dfy, dz, in the expression (1) for f^>, be such that
dx fh/ dz , .
d_Y_dZ^(lz'_t(X~clX_dY ^''
dz dy dx dz dy dx
then dp will be the variation of p as we pass from one point to the
adjacent point of any of the curves of which these are the dif-
ferential equations. Now, combinmg (l) and (3), we get by (2)
dp = 0,
wherefore p is constant along the curves whose differential
equations are (3).
Also from equations (A) we may put equation (3) in the fonn
dx dy dz
dy dz dz dx dx dy
dy =. r(x'^ - Zf
\ dz dx
dz = r{Y^i-xf\:
V dx dyj
and multiplying these equations by y- , y- , -^ , respectively,
we get ' ^
dp = -fdx + -^ dy + ~dz = 0:
dx dy dz
or p is also constant along these curves.
HYDROSTATICS, 327
1850.
1. Three equal cylinders arc placed in contact upon a
horizontal plane, sufficiently rough to prevent sliding: find
how much water must be poured into the space between the
cylinders, in order to disturb the equilibrium.
Let h be the depth of the water poured in when each
cylinder is on the pomt of turning about a tangent line to its
base, in which case the water will run out between the cylinders.
Now the moment of the fluid pressures upon each cylinder
about the tangent line to the base about which the cylinder
would begin to turn, is the same as the moment of the fluid
pressures on a vertical rectangle of height h and breadth equal
to the radius [r] of each cylinder about its base*
= /' I gp [h — z) zdz
Now this must equal the moment of the weight [Mg] of
each cylinder about the same line or Mgr^
is the required height.
2. All space being supposed filled with an elastic fluid whose
volume at a given density is known, the particles of which are
attracted to a given point by a force varying as the distance :
find the pressure on a circular disc placed with its centre at the
centre of force.
Let fi = absolute force of attraction at distance unity ; the
attractions A', Y, Z^ parallel to the axes at the point [xyz] are
• For this is the natiire of the sectiuii of each cylinder supposed of a
height /(, made by a plane through the lines of its contact with the other
cylinders.
SOLUTIONS OF SENATE-HOUSE PKOULEMS. [1850.
— /i.r, — /i^, — t^^i tlic centre of force being origin: hence
dji = p [Xdx + Ydy + Zdz)
= — fip [xdx + ydy + zdz) = — yuprdr if r'- = x^- + ?/■' + s'' ;
and p=kp\
.'. -^ — — ukrdr ;
To detennine C, wc have
p = kp = CkB-^'^"" •
BM = mass contained between two con-
secutive spheres having C for cen-
tre, radii r and r + Sr
= 47rpr'S?- = A7rCk.rh'^'""-'Br',
. 3/ = whole mass, and therefore known,
ATvCk
Let ^jfikr^ = 2,
and r'Wr = z^dz •
.-. M=4.7rCk-=— / sW^:
and
/%^s-v. = r(f) = ir(i)
and M is known ; hence G is also known.
Hence, if P be the pressure on the annulus (radius a) we have
8P = 27rrBr.j)
1850.J HYDROSTATICS. 329
from r = 0) /^'^
3. A hollow cylinder is filled with inelastic fluid and made
to revolve about a vertical axis attached to the centre of its
upper plane face with a velocity sufficient to retain it at the
same inclination to the axis. Find at what point of the face
a hole might be bored without loss of any fluid.
Let ft) be the angular velocity of rotation : then, if the fluid
were contained in an open vessel, the latus-rectum (Z) of the
generating parabola of the free surface would be -^. Now
since it is supposed by the question that there is a point in the
upper plane face where the pressure of the fluid is zero, it is
manifest that the face touches the above free smiace at this
point. This point will evidently lie in the diameter of the face
most inclined to the horizon, at a distance r suppose from the
centre of the face. Let a be the inclination of the face to the
vertical, li the distance of the vertex of the supposed free smiiaee
above the centre of the face, the equation to the free surface is
and for a;, y we may write r cos a, r sin a,
.-. r^ sin'^a = Z (r cos a — h) :
the roots of this equation are equal,
, , cosa
.-. r = \l ^-^ .
sm a
g cos a
to sm a
4. A mass of inelastic fluid is contained between three co-
ordinate planes, each of which attracts with a force which varies
330 SOLUTIONS OF SKNATE-HOUSE PROBLEMS. [1850.
as tlic distance, and the absolute forces of attraction ytt,, yu,.^, fx^^
are in harmonic progression. Half an ellipsoid is fixed with
its plane surface against one of the coordinate planes, and its
surface touching the other planes ; its axes being parallel to the
coordinate axes and proportional to //-,"*, yu,./*, fi'^. If there
be not sufficient fluid quite to cover the ellipsoid, the uncovered
part will be bounded by a semicircle.
The attractions X, F, Z^ parallel to the axes arc - yu-jir,
.'. flp = Xdx + Y(hj + Zch (if p = unity)
= - {H'.sccIx + fi_^ydy + fi^zdz) ;
therefore the equation to the free surface is
fM^x^ + fi^^y'^ + fi^z^ = a constant = C suppose (1).
The equation to the ellipsoid, if it be bisected by xz^ is
ti,[x-af + ^.y + fJi,{z-cy = C (2).
(1) — (2) gives for the plane of intersection
2fjb^ax + 2/Zg02; = a constant = 2 ( C')^ A suppose ;
••• M'l^x + /*3*^ = -4 (3),
since a — 1 , , <- — i —
(3) may be put in the fonn
fj,/' = A' - 2Afi^^x + fi^x^ ;
subtracting this equation from (1) gives
2/A^a?" + fJ^,f = 2Afiix - A^ + C (4),
the equation to the projection on {xy) of the curve of intersection.
Let <f) equal the angle at which (3) is inclined to xi/ ;
and cos"rf> = — — — = -^
since /a,, /w,^, ^u,.^, arc in hannonic progression.
tan0 = I'-i
V,
1850.] HYDUOSTATICS. 331
This equation, taken with (4), shews that tlie axes of the
projection on xij of the curve of intersection parallel to x and y
respectively, are in the ratio of cos0 : 1; hence the curve of
intersection must be circular, evidently a semicircle, whose
diameter lies in xz^ and its plane pei'peudicular to xz.
5. A rectangular vessel is filled with fluid of twice its
weight, and placed with its open end downwards upon a hori-
zontal plane, which is then made to revolve round each side
of the base successively, one of these sides being greater and
the other less than three times its height: find when the fluid
will begin to escape in each case, supposing the centres of
gravity of the vessel and the fluid to coincide.
If the vessel had a base instead of being opened at the lower
end, the moment of the fluid pressure on its inside about any
side of the base would be the same as that of its weight acting
at its centre of gravity : hence, when the vessel is open at the
lower end, the moment of the fluid pressm-es about a side of the
base will be that of the weight acting at its centre of gravity,
minus the moment of the fluid pressures on the plane on which
the vessel rests.
To find this moment, M suppose. Let the horizontal plane
be supposed to have been turned through an angle a, and let r
be the distance of any point in it from a horizontal line in it,
at the same height as the highest edge of the vessel : the dis-
tance of the edge about which the vessel is being turned will be,
if a be this edge, h the other edge, and h the height of the
vessel, h -f h cot a. Hence
M = i f/pr sin a . adr [h + h cot a — r)
J Acota
pb*hcoti
= gpa sina I [{b + h cota) r — /•"} dr
J A cota
= gpabsma{^{b+hcota) {b + 2hcota)-^[b''+3bhcota+3h''Qot''a)]
= gpab sina {\l>^ + \1d} cota).
Let Whc the weight of the vessel, and therefore 2 W that of
the fluid: then, when the water begins to flow out, the ni(»ment
332 SOLUTIONS OF SENATE-HOUSE TKOBLEMS. [1850.
about tlic edge ct of all the forees on this vessel, including its
weight, is zero, or
3 W{^b cos a — ^h sin a) — M = 0,
or, since W = gpabh, and b = Snh suppose,
— [Stik cosa — h sina) — S7ih sina {^.3nh + ^h cota) = 0,
or 3« cosa — sina — n {n sina + cosa) = 0 ;
2n
.'. tana = -2 r;
which gives the value of a when the two values of n are
substituted, one >, the other < 1. In both cases, however,
a is < 45°.
( 333 )
HYDRODYNAMICS.
1848.
1. A cylindrical vessel, with its axis vertical, is filled with
fluid, which issues from a great number of small orifices pierced
in the side : find the surface which touches all the streams of
spouting fluid.
This surface is evidently a surface of revolution, having the
axis of the cylinder for axis. Its generating curve is the line
which touches all the parabolic jets of water from the different
orifices in the same generating line of the cylinder. These jets
have all this generating line for axis, and a common directrix
in the plane in which they lie, viz. the horizontal line at the
level of the surface : for the velocity of efflux is that due to the
distance from this line.
Hence, making the common axis and directrix axes of x
and y respectively, the equation to the jet whose point of efliux
is at a depth h is
f = Ah{x-h).
To find the line which this cuiwc always touches, differentiate
with respect to /*, considering x^ y constant ;
.-. 0 = a; - 27*,
and eliminating ^,
f = 'lx.\x,
or y = x\
the equation a straight line through the origin, inclined to the
vertical at an angle of 45°. Hence the surface required is a
right-angled cone placed on the cylinder in an inverted position.
2. A closed vessel is filled with water, containing in it a
piece of cork which is free to move ; if the vessel be suddenly
334 soi.rTioNs uF senate-house prohlems. [1849.
moved forwards by a blow, shew that the eork will shoot for-
wards relatively to the water.
Suppose, for an instant, the eork removed, and Its plaee
occupied by solidified water ; when the blow is stnick this mass
of solidified water will instantaneously receive a velocity V equal
to that of the surrounding water, and the impulse on it will be
MV^ if M be its mass. But when the cork is in the place of
this solidified water, the impulsive actions on it of the suiTOund-
ing' fluid will be the same as they were on the solidified water,
and therefore the impulse on it will be the same. But the cork
is lighter than the same volume of solidified water, and therefore
the same impulse will impart a greater velocity to it, or the cork
will move forward relatively to the water.
3. A closed vessel is filled with water which is at rest, and
the vessel is then moved in any mamier: apply the principle
of the consei'vation of areas to prove that, if the vessel have any
motion of rotation, no finite portion of the w^ater can remain
at rest relatively to the vessel.
The principle of consei'vation of areas about any axis must
apply to the whole mass of water. But if any portion of the
water remain at rest relatively to the vessel, we may suppose
it to become solidified and rigidly attached to the vessel without
altering the motion of any particle of the water : but in this
case it is evident that the principle of the couseixation of areas
about any axis must also apply to the part of the -water not
solidified ; consequently it must also apply to the solidified poi-
tion of the water which, since the water is originally at rest,
can therefore have no motion of rotation, which is absurd if.
the vessel have any motion of rotation. Therefore, if the vessel
have any motion of rotation there cannot be any finite portion
of the water which remains at rest relatively to it.
1849.
1. Supposing the effect of friction in the case of aerial
vibrations in a tube of uniform bore to be the production of
1849.] HYDRODYNAMICS. 336
a retarding force on each particle equal to / x velocity,
prove that the equation of motion will be satisfied by taking
cs'^^' sm — ■{ a
f 1 — Y— ^ j t — xl as the type of the vibrations.
Let X be the coordinate of any particle at rest, oc + ^ its
coordinate when displaced at time t ; then the equation of motion
will be
df ~ clx' J dt '
Now, if we assume
P = cr^^' sin —- (nat — x), where 7?'^ = 1 — '. ., ., ,
^ \ ^ '^' 16(1-77 '
we have
-~ +/f = cz'^-^' ■! — T— cos — {7iat - ^) + ^ sin — {nat — x)[',
,„ 47r''^fl^ . 27r . ,
= — c£ -•" ■; sm — (wa^ — ic),
by substitution of the value of n.
Ai '2 ^"1 47r'''(t''^ ,,, . 27r , ,
Also a -T-T, = ^-5- «?£ *-^^ sm ^{nat — x)\
dx- X X ^ '
' ' df '^•^ dt dx'
^^ d''^ d"^ d^
and the equation of motion is satisfied.
2. Steam is nishlng from a boiler through a conical pipe,
the diameters of the extremities of which are D and d respec-
tively : prove that if V and v be the coiTesponding velocities
of the steam,
r = T — £ =*
336 SOLUTIONS OF SEXATE-IIOUSE PROBLEMS. [1851.
where k is the pressure divided by the density, and supposed
constant. The motion may be supposed to be that of a fluid
diverging from a centre, the centre being the vertex of the cone,
of whicli tlic pipe fonns a portion.
Let J) and p be the pressure and density at the distance ?•
from the centre of motion at the time t, when the velocity at
that point is u ; then, since the motion is wholly radial, its
equation is
1 dp du du . . . . - , .
~ 'j~ — ~ ~ji ~ ~ '^zr y smce the motion is steady ... (1).
Also 2) = hp (2).
The consideration of continuity gives the equation
?//)r^ = constant,
or ^ipr^ = constant (3).
From (1) and (2),
k dp du
p dr dr ^
or k \o^p = C — ^ii\
Let P, p be the pressures at the two extremities of the pipe,
But from (3),
^•log
P
w-
or
P _
P ~
£ 2* .
P
vd'
P~
VB''
. V =
' d'
185L
L If a regular homogeneous tetrahedron be placed in any
position whatever in a fluid whose density varies as the depth,
shew that when the resistance of the fluid is neglected, the
1851.] HYDRODYNAMICS. 337
7A4
tetrahedron will make vertical oscillations in the time 2-77
h being the depth of the centre of gravity of the tetrahedron in
the position of equilibrimn.
We shall first shew that the centres of gravity of the tetra-
hedron and of the displaced fluid are in the same vertical, what-
ever position the tetrahedron occupies in the fluid.
Let the centre of gravity of the tetrahedron be at a depth z
below the sui*face, and be taken as origin of rectangular co-
ordinates [xyz)^ the latter vertically downwards. Let the den-
sity at a depth z below the surface be fiz : the density at the
point xyz will be ix[z' + z) = c + fiz suppose.
Let r, y, i, be the coordinates of the centre of gravity of the
displaced fluid ;
.-. (mass) X = JfJ{c + fiz) xdxdydz :
but
JJJxdxdydz = 0,
because the centre of gravity of the solid is origin, and
JJfxzdxdydz = 0,
because every system of rectangular axes through the centre of
gravity of a regular solid is 2i 2)'>'incipal system ;
.'. ^ = 0, and similarly y = 0 :
hence the centre of gravity of the fluid displaced lies in the
vertical line through that of the tetrahedron.
Thus, in whatever position the tetrahedi'on be originally
placed, its centre of gravity will move m a vertical line, and
make finite oscillations in that line.
The force acting downwards on the solid at any time
= the weight of the solid — weight of fluid displaced
= g(TV -g fjj{c + fiz) dx dy dz
(if V be the volume of the tetrahedi'on, o- its density)
= go- V — gc F,
J5.
""' -df+''^'=^^
338 SOLUTIONS OF SENATE-HOUSE rilOBLEMS. [1851.
since the centre of gravity is the origin of coordinates, and
therefore
fjjz dx dy dz = 0.
Hence the equation of oscillating motion is
since c = /xz'j
and the time of an oscillation
Now A, the depth of the centre of gravity in the position, is
d'^z
the value of z in the above equation, when —p^ = 0 ;
or A = — ,
and the time of an oscillation
This proposition is equally true of any homogeneous regular
solid.
( 339
GEOMETRICAI. OPTICS.
1848.
1. Compare the brightuess of the Earth as seen from Venus
with the brightness of Venus as seen from the Earth, supposing
the sizes and reflecting powers of the two bodies equal.
Let Sy E, Vy (fig. Ill) be the respective positions of the Sun,
and the centres of the Earth and Venus, at the time when their
brightness is to be compared.
From E di-aw the straight lines JSa, Eb perpendiciilar to ES
and EV m the plane of the ecliptic, and similarly T c, Vd per-
pendicular to VS and VE: the part of the Earth seen from
Venus will be contained between planes pei'pendicular to the
ecliptic through Ea^ Eb ; and the part of Venus seen from the
Earth between the planes Vc^ Vd.
Let Q be the quantity of light that falls upon a imit of
the sm*face of Venus which has the Sun in its zenith. To find
the quantity of light reflected to the Earth from ajiy element
8S of the sm-face of Venus.
Let the latitude and longitude of the element B8, referred
to the Sun as origin, and plane of the ecliptic as plane of
longitude, be 6 and (f). The quantity of light reflected to the
Earth from SS will
= QBS X cosine z. d. of Sun x cosine z. D. of Earth
= QBS.cos9cos(li.cosecos{V-(f>), F= l SVE,
= Q . r'^ Bd cos^ B(f) . cos^O cos<f) cos( V— ^),
if r = radius of the Earth or Venus.
Hence the whole light reflected to the Earth
= Qr'JJcoB'd cos</) cos(r-<^) dd d<f)
= i<3r7/cos''^(co3F+cos(F-2<^)} dd d(f>
= ^Qr'Jcofi'e{cosV.<f>-^sm{V- 2<^) + Oj <7^:
z2
340 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
from <f> = {V— ^7r]]
to </) = Itt
= ^ Qr"" Jcos'e [cos r (tt - F) + ^ sin (tt - T^)
= ^ Qr-' {(tt - F) cos V+ sin F} /cos'(9 tZ^
= ^Qr^ {(tt- F) cosF+ BinF}/(cos3^ + 3 cos^) d0:
from ^ = — ^tt"!
to 0 = + i-TTJ
= |(2^.2((7r-F)cosF+sinP"|.
Similarly, the whole light reflected from the Earth to Venus
= f Q'r' {(tt - E) cosE+ sin^},
when Q' is the quantity of light that falls upon a unit of the
Earth's sui'face which has the Sun in its zenith ;
and the required ratio
_8V^ (tt-E) cosE+s'mE
~ SE' ' (tt - F) cos F+ sm F •
2. Find the geometrical focus of a pencil of rays refracted
through a hollow glass sphere, whose external and internal radii
are ?•, r' re^ectively.
Let ti, Vj, ^25 ^3) ^^^ ^3 be the distances from the centre of
the sphere of the foci before the 1^' and after the P', 2°^*, 3'^'^, and
4"^'' refractions respectively.
Then, by the common fonnula,
i = _^i:zi + ^ (,),
1- -'LLI. 1 i
1848.1 GEOMETRICAL OPTICS. 341
or t^tzl + 1. (5),
V, r v^
i=^+^ (3),
d ^ At" ^ 1 1
or :^ = - C + _ (4 .
Adding (1), (2), (3), and (4),
1 1\ . At
and for glass, /i = f ;
1 Q n i\ 1
which determines v,.
V, 3 U' rj "^ w '
3. Light, proceeding from a given point P, suffers any num-
ber of reflections and refractions : if consecutive rays of a given
colour come out parallel, in the direction determined by angular
coordinates ^, (f), shew that -y- , -^ may be obtained by dif-
ferentiating as if the differently coloured rays which severally
come out parallel to their consecutives started from P in the
same direction.
Application. In the case of the rainbow of the 2^^^ order,
given
D = pjT + 2(f) — 2{2)-\- 1) (^', sin^ = /i sin0',
find the order of the colours.
In general, if ^, ^ be the coordinates upon cniorgencc of the
ray whose coordinates as it proceeded from F were 0\ <f>\ and
if /A be the refractive index of the ray,
342 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
Now, If 0 and (f) be the coordinates of the rays of refractive
Index fji whicli come out parallel to their consecutives, 0', <f>'
may be foiuid in terms of /x. from the equations
d6' '^ d<f>" dd' ~ '
d^ d^ ^' _
dS' dii>' ' dB' '
those consecutive rays being supposed to come out parallel
whicli before Incidence lie In a plane defined by the particular
value of -^ .
du
Now, supposing 0 and </> to retain the particular meaning
assigned to them above, since & and <^' are now fimctlons of /t,
dB (dd dO d<f>'\ dO' dO
dfi [dO'"^ dcf>"de') dfi'^ dfi'
d(f> _ /d(f) d(f) d<}>'\ dO' #
dfi \dd' d(f)' ' dd' J dfi dfx '
The first tenns In these expressions for -7- , -^ correspond
to the variation of the direction of emergence due to the va-
riation of the direction of Incidence ; the second tenns coiTcspond
to the variation of the direction of emergence due to the va-
riation of /A In the differently coloured rays. But we have seen
that the first tenns are each equal to zero ; hence the whole
variation of 6 and <^ is due to the vaiiatlon of fj, In the dif-
ferently coloured rays, or we may obtain -7- , ■— by differen-
tiating as if the differently coloured rays which severally come
out parallel to their consecutives started from P in the same
direction.
In the application to the case of the rainbow all the incident
rays are parallel ; we may, however, differentiate for -^ , -^
as if all the angles of incidence of rays which come out parallel
to their consecutives were equal.
1848.] GEOMETRICAL OPTICS. 343
Here each ray which comes to the eye moves in the same
plane; and since the rays incident upon the raindrop are all
parallel, the angular coordinate after emergence will be i),
the deviation. Hence, to find -y- we differentiate i), consider-
dD „, ^, d<i>'
ing 0 constant;
and s,in<f) = fi sin0' ;
... ,, dd)
.'. 0 = 8m9 + /J, C0S9 -J- )
and -J— = —^ tan0 is positive :
hence the red rays which come out parallel to their consecutives
will be more deviated than the violet rays which come out
parallel to their consecutives.
It only remains to find in which direction the rays which
form the rainbow have been deviated. To ascertain this, we
must differentiate D with respect to </>, considering fi constant,
and put
dD ^
d^ = '-
This equation will give us a value of ^, which substituted in D
will determine the amount of deviation of the rays of the re-
fractive index /t, by which the corresponding part of the rainbow
is seen : let this value be
D = 2mir -f i/r :
then, if a^ be < tt, the deviation at the first refraction will be
towards the eye, and the red rays will appear on the inside of
the arch : if i/r be > tt, the deviation at the first refraction will
be from the eye, and the red rays will appear on the outside of
the arch.
4. Every diameter {d) of the extreme boundary of a sphe-
rical reflector subtends a right angle at C the centre of the
sphere: supposing parallel rays (inclined at an angle a to the
^44 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
axis of the reflector) to be incident upon every point of the
extreme boundary, shew that the section of the reflected pencil
made by a plane passing through C and peqiendicular to the
axis, will be an hyperbola, whose axes are d and J cot a.
Let CO (fig. 112) be the axis of the mirror AOB-^ let Or,
parallel to the direction of incident rays, be taken for axis of x ;
Cy, pei-pendicular to it in the plane OCx^ for axis of y, and Cz
pei-j^endicular to Cb, Cy for axis of z.
Let Cx pierce the miiTor at the point o: Join P, any point
in the bomidaiy of the miiTor by arcs of great circles, with
0, o : call Po, PoA, 0 and </> : OF will be 45°.
The ray reflected from P will pass through the point D
of Co, such that the perpendicular from D upon CF bisects CP,
t. e. through the point {^a sec 6^ 0, o) : also the coordinates of P
are a cos^, a smd cos^, a ski 6 s'm(f>. Hence the equation of
the reflected ray is
X — ia sec6 ii z , .
a 1 a = ^a ± = — • a • — T = ^ SUppOSe...(l).
aeost/ — ^a secc^ asmc^cosip asmc^sm9 ^^ ^
Also, from the triangle OPo,
cos45° = cos^ cosa — sin^ sina cos(^ (2).
Our object will be to eliminate ^, <^ from these equations,
and find the relation between r\ and z when we have written.
a; = ?; sin a,
y = 7] cosar
the equation so fonned will evidently be the equation of the
curve in which the plane through C cuts the surface formed by
the refracted rays.
From (1),
X cos 0 — la = a\ cos^d — \a\ ;
.-. aV cos'a - a\x cosB + ^x^ = ^a^X{\ - 1) + Ix' ;
.-. a\ cos e = ^x+ {^d'\{\ - 1) + lx'}i (3)-
Hence equation (2) becomes, multiplying by aX,
-^a\ = [^x + {^a^X{X- 1) + ^x^] cosa - y sina,
1849.] GEOMETRICAL OPTICS. 345
or {|rt'''\(X— 1) + ^x"]* cosa = -^ a\ + y sina — ^x cosa,
it"
or {^a''X(\— 1) + \7f sin* a]* cosa = ^ «^ + ^ sina cosa ...(4) :
hence, squaring,
\ci\ (A. — 1) cos'^a = \d'''}^ + -^ ciXt] sina cosa ;
.'. aX sin'''a + a cos'^'a = — %^i) sina cosa,
and aX = — 2-7; cot a — a cot'^a.
Again, from equations (1),
d'X' Qm^e = f + z' :
adding this equation to (3)'^,
«V = ^x' + f + z' + K^(X- 1) + {ia'X(X- 1) + \x']^-x^
or by (4),
ia'^X (X + 1 ) = ^77^ + i77^cos'"'a + z^ ^ ( tu: «^ + \'r] sina cosa ) « sina ;
cosa \2* y '
or retaining only the second powers of t] and z^
Iff cot'a = \'i]\\ + cos'a) + z"* - tf + ^t^'' sin'a + &c.,
or T\' cot'^a — z'' = &c.;
shewing that the required locus is a hyperbola, the ratio of
whose axes is cot a.
Now the axis in the plane OCx is evidently d: hence the
axes are d and d cot a.
1849.
1. A ray of light is incident upon one of two reflectors
inclined to each other at an angle - , in a direction parallel
to a line which is at right angles to their intersection, and
bisects the angle between them : supposing the intensity of a
ray reflected at an angle (j> to be to that of the incident ray
as e cos(f> to 1, shew that the intensity of the ray after it has
suffered n reflexions will be to that of the incident ray as e"
to 2"-'.
346 SOLUTIONS OF SENATE-HOUSE PROBLEAIS. [1849.
From the problem on p. 60 it appears that, if n be even, the
successive angles of reflexion are complcmentaiy of
Itt S tt 7} — 1 it n — 1 tt Stt Itt
2'w' 2'n '" 2 'w' ~~2 'n"'2'n^ 2n'
Ilcncc the intensity of the ray after n reflexions : that of the
incident ray
-"(
. 1 TT . 3 TT .71—1 TtV ,
sm - - . sm - - ... sm — - — . - : 1.
2 n 2 71 2 nj
Similarly, when n is odd, this ratio is
„ / . 1 TT . 3 TT . n ttV .
e sm - — . sm - - . . . sm - . - : 1.
V 2 w 2 n 2 nJ
Both these ratios may be expressed by the general formula
„.l7r .37r .2?i — Itt,
e siij - - . sm - - . . . sin — : 1 :
2 71 2 n 2 w '
and in Hymers' Tlieory of Equations^ Art. 22, Ex. 20, it appears,
by making ^ = 0, that
. 1 TT . 3 TT . 2« - 1 TT 1
sm - - sm - - . . . sm — - — - = -^r=-, ,
2 n 2 71 2 w 2 "
whether n be odd or even : hence each of the above ratios
= e" : 2"-\
2. A transparent medimn is bounded by two parallel planes ;
the refractive index is constant thi^oughout any plane parallel
to the bounding planes, but varies continuously in the direction
of the nonual to those planes: shew how to find the path of
a ray of light tlu'ough such a medium, and prove that in passing
through a section for which the refractive index is a maximmn
or a mmimmn, the path will in general have a point of contraiy
flexure.
It is evident that the path of any ray will lie in one plane :
in this plane take two lines, one perpendicular to the bounding
planes, the other parallel to them as axes of x and y re-
spectively.
1850.] GEOMETRICAL OPTICS. 347
At the point [xy] let the inchnation of the path to the axis
of X be 0, and let the refractive index be yit: at an adjacent
point [x + Sa-, ?/ + Sy), let these be ^ + 8^, fi-\- Bfi: then, by the
law of refraction,
Bin<f> = — sm{(f> + S(f))
+ — 1 (sin^ + COS080),
or 0 = cos ^B(f) ■+ — sin</)S/i;
therefore, proceeding to the limit,
-J- + - tan<^ = 0,
, , d6 1 da
or cot<i> ~ + - -f- = 0'.
ax fi ax
.'. logging + log/i = constant = logO,
.*. sin© = — ,
01* 1 + I ^- I — I Vy / )
dxV _ ffM
dy) " \G
dy f//.^^ ^-*
which, since /i is a known function of a*, is a diflferential equation
for the determination of the path.
7
When /x. is a maximmn or minimum, we have -y- = 0,
and as the ray passes through such a section, its path usually
suffers inflexion.
1850.
1. If a string be wrapped round a glass prism, whose section
is an equilateral triangle, so as to be always inclined at the same
348 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
angle to the axis of the prism, the portion of the string seen by
internal reflexion will appear to be parallel to the portions seen
directly.
Let AB (fig. 113) be a portion of the string seen directly,
BC a portion seen by internal reflexion at the surface AC oi the
prism : and first, let the eye be In such a position that the small
pencil of rays by which any point of BG very near B is seen,
shall pass through some point of the surface ABB' very near
to B\ a point in the edge AB'. Then, if the eye be at a con-
siderable distance from the prism so that the axes of small visual
pencils may be considered parallel to each other, any point of
BC very near C will be seen by a small pencil which passes
through the surface ABB' at some point very near C a point
in the edge BC. Let B'U^ G'E be the directions of these small
pencils upon emergence. Now the plane CC'E is parallel to
the plane BB'E] and if we draw a plane through A parallel to
these, the plane BB'E will be equidistant from the other two,
since the prism is equilateral, and AB^ BC equally inclined to
its axis; hence BC = AB'^ and B'C is parallel to AB:
hence BC appears parallel to AB.
Now let the eye be moved in any manner without approaching
too near the prism ; it may easily be seen that the locus of the
points where the rays from BC to the eye cross the plane ABB'
is parallel to B' C", and therefore to AB. Hence BC will always
appear parallel to AB; and the same may be shewn of any
other portion of the string seen by internal reflexion.
2. A rectangular box, at the bottom of which is a plane
mirror, contains an unknown quantity of water ; from the angle
at which a ray of light must enter through one of two small
holes in the lid in order that after refraction and reflexion It
may emerge at the other, determine the height of the- water in
the box.
Let a be the height of the box, x the depth of the water,
2h the distance between the small holes in the lid j ^, (f)' the
angles of Incidence and refraction when the ray enters the water.
1850.J GEOMETRICAL Ol'TICS. 349
they will also be the angle of refraction and incidence as it
emerges from it : hence we have
h = [a — x) tan 0 + ic tan 0',
sin0 = /isin^';
a tan 6 — h
tan^ — tan^'
a tan (j) — b
, sin cb '
tan</)-7-^ ^^---1
(/Lt — sni 9)*
whence x is known from <^, which is the angle the ray makes
with the vertical upon entering the hole in the lid.
3. If a kiminous point be reflected by a small plane mirror,
so as to be seen by an eye in a given position, and the mirror
move in such a way that the Imninous point always appears to
be upon a given conical surface, of which the point is the vertex
and a line through the eye the axis ; find the form of the sm'face
upon which the small mirror must always be situated.
Let 0, E (fig. 114) be the position of the luminous point and
eye respectively, 31 any position of the miiTor, P the corre-
sponding position of the image of 0 : then will EMF be a
straight line, and MP = MO. Take OEx for axis of x, Oy
perpendicular to it for that of y ; x^ y the coordinates of M\
x\ y' those of P. Then
MP = MO ;
.-. {x'-xy+{y'-yY = x' + y%
or x' + y"' - 2xx' - 2yy' = 0 (1):
JJ, M^ P, are m the same straight line,
.-. -y-^J— (2).
X — a X — a ^
Let the equation to the generating line of the cone on which
P is situated, be
y = mx (3) :
between (1), 2), (3), we have to eliminate x', y.
350 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
From (2) and (3),
x - a _ m[x — a) ^
X y
y — ra[x — a) '
From (1) and (3),
(1 + Q)f) x — 2[x + my) = 0,
or 2{x + my) [y — m[x — a)] — (1 + vi^) a?/ = 0 ;
.-. 2m{y'' — x'] + 2{l— m') xy + 2'tnax — (1 - m') ay = 0,
the equation to an equilateral hyperbola; and the required
surface is an equilateral hyperboloid of revolution.
4. If the earth, supposed spherical, were covered to a depth
h with water, h being small compared with the earth's radius,
shew that the height to which a person must be raised above the
surface of the water in order to see as far below the horizon as
when he was on the surface of the earth is ^ , .. r nearly,
2r{fM'- 1) •^'
fi being the index of refraction for water.
Let 0 (fig. 115) be the centre of the earth, A the station of
the observer on the earth's surface : in order to see as far below
the horizon as possible he must look in the direction AB, such
that OB A is the critical angle ; he will then see objects situated
in the line 5(7, if i? (7 be a tangent to the surface of the water
at B. Hence P, the raised position of the observer, must be the
intersection of CB, OA. Let A OB = 6 : then, if AP = x,
or, since ^, - , - , are small
r +
r +
h _
X
= cos^.
lall,
X -
r
-h
=1-
X -
-k
= ie'.
1-— -1=1-^6^;
1851.] GEOMETRICAL OPTICS. 351
Also from the triangle OAB^
■ / • -1 1
, sin sni — +
r + h \ /x
• • -il
sin sin -
(A
h
or l+^=l + (/.'^-l)i^,
or ^ = -—a TTT— 3
or x — Ti —
2r' {fjC' - 1) '
2r (/Lt' - 1) '
the required distance to which the man must be raised above the
surface of the water.
1851.
A number of vertical plane reflectors are placed together so
as to meet a horizontal plane in a polygon of n sides : find the
path of a ray of light which, after reflexion at the n plane
reflectors in succession, will continue to proceed in its onginal
course.
Shew also that when there are four reflectors the problem
is either indeterminate or impossible ; and that when the number
of reflectors is even, and the polygon capable of being inscribed
in a circle, the problem is indeteraiinate.
Let ^j, O,y..0^ be the complements of the successive angles
of incidence or reflexion, a^, a,^...a„ the angles of the polygon:
then ^j, ^.^, flj, are the angles of a triangle ;
so 0„ + 0^ = TT — a
.(A).
0, + ^, = TT
352 SOLUTIONS OF SEXATE-IIOUSE PROBLEMS. [1851.
Again, let ?/, = 0, u^ = 0,... u^ = 0, be the equations of the
successive parts of the ray's path, a^ = 0, a^ = 0,... a^ = 0, the
equations of the sides of the polygon ; the equations being all in
such a fonn that u considered as a function of x and y is the
distance of the point [xy) from the line u = 0.
Now Oj is the external bisector of w^, w^, whence
Wj + u^ = 2a^ cos 0^.
For let Owj, Oti^^ Oa^^ (fig. 116) be the lines w.,, Wv,, o, ; take
any point P, join OP, and draw P>-, Ps^ Pt, perpendicular to
these lines respectively. Then
Pr + Pt= OP (sill POr + sin POf)
= 2 OPsin^iPOr + POf) co^{POr - POt)
= 2 0PsmPOs costOs
= 2Pscos6^.
Hence, if Xj y, the coordinates of P, be substituted in Mj, u^^ a^,
we shall have
u^ + u^ = 2a^ cos^j,
and the same may be shewn wherever the point P is taken ;
hence generally,
u^ 4- ti^ = 2a^ cos 0^:'\
so u^ + u^ = 2a^cose^ \
■?/, 4- u = 2a cos^
» ' 1 n n J
From equations (A) 6^^ ^.2V ^„ must be found, and thence
Wj, «*2V ^n from equations (B), and the path of the ray will then
be fully determined.
If there be four reflectors we find, by adding the P' and S'^
of equations (A),
O, + e^ + 0, + 0, = 27r-a^-^,:
.similarly, by adding the 2"<* and 4}^^
^, + ^. + ^3 + ^4 = 2vr - a.^ - a^ :
1851.] GEOMETRICAL OPTICS. 353
hence we must have
a, + a, = a^ -f- a^ = TT,
or the quadrilateral must be inscrlbable in a circle in order that
the problem may be possible.
If however this condition is satisfied, still the problem is
indetemiinatc ; for if wc treat equations (B) as above, we find
ttj cos^j + rtgcos^j = a^ cos6^ + a^cos^^ (1).
This is an identical equation ; It will therefore lead us, by means
of equating the coefficicjits of x aiid i/ on the two sides of the
equation, to three conditions between ^,, 6,^^ 6^, 6^^ and constants :
between these constants there ai'e also two relations arising from
the circumstance that the quadrilateral may be inscribed in a
circle, and the three conditions between ^,, 6^^ 6^^ 6^^ and
constants amount to only one equation independent of equa-
tions (A).
This condition, together with equations (A), will detci*mine
^1? ^2i ^35 ^4- ^^^ since we have derived the condition (1) from
equation (B), it shews that those equations are equivalent to
only three independent equations, and ii^^ u^, u^, u^ are therefore
indetemiinatc. The dii-ection only of the different parts of the
path of the ray can be determined; with these directions any
position will satisfy the problem.
Similarly, if there be any even number of reflectors, we may,
from equations (A), deduce the condition
a, + 0(3 + ... + a„_, = (x^ + a^ + ... + a„
or the siuns of the alternate angles must be equal : this condition
is satisfied if the polygon can be inscribed in a circle, and the
problem is then possible.
The problem is still indeterminate, for from equations (B)
we may deduce the condition
a,cos^,+a3Cos^34-...+ a„_,cos^,,_,=rtjjCos^2+a^cos^_, +...+ «„co8^,, :
this Identical equation will lead, as before, to one equation of
condition independent of equations (A) between 6^^ 0^, ... 0^^
and constants, which, with equations (A), serves to detennine
AA
354 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
O^^O.^y.. 6^. Still equations (B) are equivalent to only n — \
independent equations, and ?ij, ?<2, ... w„ are therefore inde-
terminable. Ill fact, we have shewn in Problem 2, page 59,
that ' if a ray of light, after being reflected any number of times
in one plane, at any number of plane surfaces, return on its
fonner course, the same will be time of any ray parallel to the
former which is reflected at the same surfaces in the same
order, provided the number of reflexions be even.'
( 355 )
ASTRONOMY.
1848.
1. If there had been uo stars, how might the absoUite
periodic times of the Earth and planets have been determined,
even if the eqnator had coincided with the ecliptic ?
We might first have detenuined the synodic time ( T) of the
Earth and any superior planet by observing the interval between
successive conjunctions. Let jE, P be the periodic times of the
Earth and the planet,
27r 27r , . , , . .
-'. 'Y ~ 'p ~ relative angular velocity of
the Earth and planet
- yr,
or E=t{i-^
We might then have observed the elongation from the Sun
of P and the other planets at their points of station : this gives
the ratio of the distances of the Earth and each planet from the
Sun, and therefore, by Kepler's law that the squares of the
periods are as the cubes of the mean distances, it would give
E
approximately the ratio of the periods or -p for each of the
planets; whence from above E^ and the periods of all the
planets, would be known.
2. A star map is laid down on the gnomonic projection, the
plane of projection being parallel to the equator: give a gra-
phical solution of the problem, to determine the time at a
AA2
35G SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
kno\NTi place l>v observing wlicn two stars laid down in the map
arc in the same vertical plane.
Since the place is known, we may draw about the centre of
the map the circle described by the projection of the place on
accomit of the Earth's daily rotation. Let a line through the
two stars intersect this circle in two points A and B. Then it
is plain from the method of projection, that when the projection
of the place is at A or B, the two stars arc in the same
vertical.
Let the differences of longitude of the Sim in its place
among the stars on the day of observation, and the points A
and B, be obseiwed ; this longitude, converted into hours at the
rate of 15 degrees to an hour, w^ill give the interval since last
noon or the true solar time.
3. Shew that at the equinoxes the extremity of the shadow
of the style of a vertical south dial will trace upon the dial-plate
a horizontal straight line at a distance acosec? from the upper
extremity of the style, a being the length of the style and / the
latitude of the place.
On the day of the equinox the Sun appears to move in a
great circle of the celestial sphere. We may consider the
extremity of the style as the centre of that sphere, or that the
Sun moves in a plane through the extremity. The nonual to
this plane lies in the vertical plane thi'ough the north and south
points, therefore its intersection with the dial-plate will be a
horizontal straight line ; this is the line traced out by the ex-
tremity of the shadow of the style.
Let the plane of the paper be the vertical plane containing
the style AB (fig. 117). Draw AG vertical and BC perpen-
dicular to AB: then, since AB is parallel to the Earth's
axis, and the Sxm is in the equator, C is the extremity of the
shadow at noon, and ^ 0 is the distance of the horizontal line
from A : it = ABcosecACB = a co&ecl.
ASTRONOMY. 357
1850.
1. If a rod be fixed into a vertical wall which faces the
south and the shadow of it be cast upon the wall by the Sun,
find the curve upon which the shadow of the end of the rod will
be situated every day at mean noon, the Sun being supposed to
move imifonnly in the ecliptic with his mean motion.
The mean Smi is situated on the equator at the same distance
from "Y* as the true Sun ; it is mean noon when this mean Sim is
due south.
Let S, S' (fig. 118) be the true and mean Suns, then
nr >S' = "V >S", and if we draw SD an arc of a great circle perpen-
dicular to the equator, and call 'Y' S, tp D, SD^ i, a, S, re-
spectively, we have
S'I)=yS-'rD = L-a.
Now let E be the extremity of the rod, 0 its shadow when
the Sun is in v , P the position of its shadow when the Sun is
at yS, and S' on the meridian, i.e. due south. Then if we draw
ON horizontal, NF vertical in the plane of the wall, and join
EO, EN, EP] OEN== S'D, NEP = SB, EPN=l - S, where
I = latitude of the place. Call On, x, JVP, y, EO, d,
.'. X = dtoxiS'd,
^{d' + xy-smSD
,, , -r . , tanZ — tana
or X = dtaiiiL — a) = d — — ; — -^
^ ' I + tana tsuiL
, tana — cos w tana . ,
= d 5 , smce cosw taniy = tana
cos<w + tan a
, (1 — cosci') tana , .
= a — f — 5 — (1) ;
coscD + tan a ^ ' '
and 3/ = {d' + xr~ -^f., = [cl' + ^y . J \ -,
^ ^ sm(/— o) ' sm^cotd — cosZ
= [d'^+x;')^ -r-5 — J—. — , since sina cotS=cot&>...(2).
sm/cotw— cos/sma ^ ''
"We have now to eliminate a between (1) and (2).
358 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1850.
From (2)
sin 7 cot G) , (d'^ + x'^)^
— ; cost = -^^ ,
sma y
• 7 V
or sin a = sin 6 cotco -r^. rrd-
{(.r + xy + ycosl'
From (1)
(coso) cos^rt + sin^a) x = d{l — cosw) sina cosa,
or (coso) + (1 — COSO)) sm'aYaf = (f (1 — cosw)'' sin^a (1 - sin^a) ;
.-. [cosw {{d^ + xy+ycosl]'^ + (1 — cosw) sin'"*? cot^ co.^^]'"' x^
= (1 - COSO))'' cof 0) sin^? d'f [{{d^ + x^ + y cos?]'' - sm7 cof eo.?/''],
the equation to the required cm've.
2. Suppose that dm'ing the day of the equinox, a man walks
in a horizontal plane towards the Sim at a miiform rate ; prove
that the equation of the path described by hun is
ny \ _ sin? ^^ ^^^^^ _^ ^ „,,^,
sin -^ + 0=^ £«'-' + e ,,
Vasecf / 2 V /
where x and y are the coordinates of his position at any time,
measured along and at right angles to his meridian at noon ;
I is his latitude, and a is the space he walks over while the
Earth revolves through an angle n.
Deduce the particular cases of his being at the pole and at
the equator.
Let 0) be the angular velocity of rotation of the Earth about
its axis ; the angle apparently described by the Sun in the time
t will be ft)^, since the Sim is in the equator, it being the day
of the equinox.
If a be the Sun's azimuth at time <,
dy
-~ = tana.
ax
Now I and mf are the sides of a right-angled triangle, sup-
posing the man to start at noon, of which the angle opposite
tot is a y
1850.] ASTRONOMY. 359
t&ncot
.•. tana = -v— ,- ,
dy tanwf
dx sin? '
■V dy _ tanw^
*°^ ds ~ (sm^Z + tau'^a)«)i '
cfe sinZ
(/5 (sin' ? + tan' «»«)**
Also, * = - G)^
n . n
- tan -s sin - s
ay a a
^ fsin^Z + tan'-sV fsin'^Z cos' - s + 1 - cos'' - sV
V a J \ a a J
. n . n
sua — s sm - s
a , a
— sec6 4 ;
[ 1 — cos''? cos''- s] f sec' ? — cos' - s j
.*. y = -secZ-jcos'M cos - scosZj — 4 ; •.• y = 0 when s = 0,
or cos -s= sec? cosf — —i+l] (1).
a \asecl J
^ sm I cos - s
. . dx a
[sm? 1 — sm -s + sm - sj
V a I a
sm? cos-«
a
( sin'' ? + cos' ? sin'' -s\
n
cos -s
7 ^
= tan?
[tan' ? + sin'' - s j
360 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
a
. w / . . . w \'
sin - s + tan"'' I + sin" - s )
a \ a )
.'. X = - tan Hog i ; '.' x = Q when s = 0 ;
n
tan /
/ \4 "'
.'. sin -s + (tan'/ + sin' - s] = tan? e° '^°' ?
a \ a J
.-. 8in-5- (tan'/+sin^-5) = - tanZ e""'^"' ,
a \ a J
or (sec'Z-cos'^sV = ^ /ga-^ + g-^ ^
the required equation to the path.
We have in this sokition considered I constant ; if, however,
the man be at the pole, I will = ^tt, and sec/, tan/ will be
susceptible of great changes when / alters but very little ; hence
we must consider his motion as indefinitely small compared
with that of the Sun, or - indefinitely small : hence the above
equation leads us to x = 0, ^ = 0 ; and the man merely stands
at the pole looking towards the Sun.
If he be on the equator, tan / = 0, and therefore a; = 0, or he
walks along the equator.
1851.
The declination of the Sun at two obsei'vations S, 8', and
the Sun's motion in right ascension and longitude in the in-
tei-val between the observations, are equal: shew that if &) be
the obliquity, and a, / the Sun's right ascension and longitude
at the first observation, cosw = cosS cos 8'; tana = sinS cot 8';
cot/ = sin 8' cotS.
Let P, K (fig. 119) be the poles of the equator and ecliptic;
>S', S' the two positions of the Sun : then, since the differences
1851.] ASTRONOMY. 361
of the Sun's longitude and right ascension at S and S' are
equal, SS'=SPS'. Let the angle P8K=^'^ draw KR^ an
arc of a great circle, to meet SP produced at right angles:
then, in the triangle SPS\
sin/S-S" ^mPSS' = sin -S'P sin /SP/S",
or cos<^ = co88'; .*. ^ = 2',
and KR = (f)=:B':
also PR = 8, KP = ft),
lKPR = 90 - a, lPKR = I
Hence, by Napier's rules,
cos 0) = cos 8 cos S' (1),
sin 8 = tana tan 8',
or tana = sin 8 cot 8' (2),.
sin 8' = cot? tan 8,
or cotZ = sin8' cot8 (3),
and (1), (2), (3), are the formulae required to be proved.
( 362 )
DISTURBED MOTION.
1848.
Two bodies, P, P (fig. 120), describe round a central body S
circular orbits lying in one plane, the orbit of P being within
that of P'; prove that the disturbing force of P' on P, when
wholly central and additions, will be equal to the disturbing
force when P, P' are on opposite sides of /S, provided SP' be
a mean proportional between SP and SP + SP.
Let Pj be the position of P when the distui'bing force [F^
on P, is wholly central and additious ;
. F -J^ ^
• > P^P"''P^P">
(where fi is the absolute force of attraction of P').
Under the condition that the force of attraction of P' on
S and Pj, perpendicular to SP^, is the same, i.e. that SPP^ is
an equilateral triangle,
.-. PP^ = PS,
IJ,.SP^
or F =
SP"
Let F^ be the disturbing force when P is in opposition to P'
at P„
P =
fi fi
if
SP" P^P" '
and P, = P,
1 1 ^P
SP" {SP^ + SPY~ SP"'
or, dropping the suffixes, because SP^ = SP^,
{SP+ spy - SP' = ^ {SP+ spy
1849.] DISTURBED MOTION. 363
or SRSP' + 2SF" = SF' + 2SRSP" + SF%
or SF'= SF{SP+SF'),
or if SF" be a mean proportional between SF and SF + SF.
1849.
If, in addition to the force of the Sun on a planet, there be
a small force tending towards the Smi, and varying invei*sely as
the m^^ power of the distance of the planet from the Smi, prove
that the perihelion of the orbit will have a progressive or re-
gressive motion, according as m is greater or less than 2.
Can you explain this result by reasoning similar to that used
in " Airy's Gravitation'''''^
If F be the whole central force on the planet we shall have
= jXU + fill ,
where fi' is very small. The equation of motion is
d% F
d'^u II u! „,_..
For a first approximation,
d^u yU-
W + " ~ P " ^'
which will be very approximately satisfied by
M = «{1 +e cos(c^ — a)],
if c be very near unity, and a = ^ ;
.-. ^^ m"'-» = ^ a"'-' {1 + {m -2) e cos(c(9 - a)|,
omitting higher powers of e.
Hence, for a second approximation,
-j^ +u- a- — a"' ' --
do fJL fl
+ u- a- — a"'"' - — a"""' [m -2) e cos {c0 - a) = 0,
364 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
which is satisfied by
u = a\l+ — a'"-' + e cos (cd - a)].
if ae{l- c') cos [cO - a) - — a'""' {m -2) e cos(c^ _ «) = 0,
A*
or l-c' = ^(m-2)a!"-':
hence c is < or > 1 according as m is > or < 2.
Now, the argument {c6 — a) may be put in the form
^-{a+(l-c)^};
whence it appears that the above equation between u and 0
is the equation to an ellipse, the longitude of whose apse is
a + (1 — c) ^; its apse will therefore progress or regress accord-
ing as c is < or > 1, i.e. according as m is > or < 2.
This result may be explained in a manner similar to that
used in Aiiy's Chavitation^ Art. 98.
Let P, A be perihelion and aphelion.
The disturbing force is towards >S' both at P and A ; it will
therefore progress about P and regress about A. To consider
which of these effects will be the greater. If the disturbing
force at P, A and the other points of the orbit were propor-
tional to the inverse square of the distance, its only eflfect would
be to alter the magnitude of the central force in a certain ratio
without altering its law ; it would therefore have no effect upon
the position of the apsides, or its eflfects about P and A would
be equal. But if the disturbing force vary inversely as the
(distance)"', where m is > 2, the ratio of its intensity at P to
its intensity at A will be greater than the ratio of the intensities
of the central force at those points ; hence its effect will be
greater at P than at -4, or the progression at P will be greater
than the regression at A ; i. e. on the whole the perihelion will
progress. Similarly, it may be shewn that if m be < 2, the
perihelion will regress.
( 365 )
ATTRACTIONS.
1848.
1. A sphere is composed of an immense number of free
particles, equally distributed, which gi'avitate to each other
without interfering: supposing the particles to have no initial
velocity, prove that the mean density about a given particle
will vaiy inversely as the cube of its distance from the centre.
The attraction upon any particle will be the same as if the
matter nearer than itself to the centre were collected there, and
attracted with a force varying inversely as the square of the
distance. This attracting mass will remain the same for the
same particle throughout the motion. Let x^ x-\- ^x be the
distances from the centre at the time ^, of two particles situated
in the same radius, whose original distances from the centre
were «, a + S« ;
d'^x fji,
•'• ~cie^~ x''
and U- =2/. --- =2^
,dtj \x aj ax
dt ( ci\^ X
dx \2fiJ (ax—x^)^
a \^ ( ^a ^a — x
2fjbJ \{ax — xy {ao; — x;y
But /A depends upon the mass originally contained within the
sphere radius a ;
.*. fi^ a^ = Ca^ suppose ;
366 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
In order to find the relation between ^u- at the time t and
8a, we must differentiate this equation, considering x and a
variable, and t constant;
_ /,, lx\ aSx — xBa
•■■ ^ -^ W a' '
or 8x = '- 8a.
a
Hence the volume of the shell originally contained between
the spheres of radii a, a + Sa, i. e. of volume iTrci^Sa^ is now of
x^
volume Attx^Sx = 47r — S« a x^. Hence the density of the
matter in this shell, Avhich varies inversely as the voliune, varies
inversely as its (radius)' : hence the proposition is time.
2. Prove geometrically, or otherwise, that if g be the attrac-
tion which a particle m exerts on a point m a closed surface 8,
0 the angle between the direction of g and the normal, doi
an element of 8^
JJg cos, Odd) = 4:77771, or = 0,
according as m is within or without 8, the attraction of m at the
ni
distance r being — ^ .
Extend this result to the case of a finite mass cut by 8, and
thence prove by taking for 8 an elementary parallelopiped, that
if V be the potential of any mass for an internal particle,
d'V d'V d'V
'd^-^W^w=-^''p'
About the particle m as centre describe a sphere of radius
unity ; and let a cone having m in in its vertex, and circum-
scribing the element dco of the smiace 8, include a portion dco'
of the surface of this sphere : then the relation between day
and dw' will be
dot) cos 6 = r\ do)'.
1848. ATTRACTIONS. 367
Also, let g be tlic attraction which m exerts on a point
at distance unity;
.■.g = l:
' r
hence g coaOdo) = g'doi'^
and fjg cosddo) = g'co'
= the whole attraction of m on w',
where w' is the whole projection of S on the surface of the
sphere: hence, if m be external to S we see, by taking the
projection of each element with its proper sign, that to' = 0 ;
but if m be within S, cu' = 47r ; and, by the question, g' = m ;
•'• !J9 cos^c?&) = 47rw, or = 0,
according as m is within S or without it.
This equation expresses the value of the sum of the attrac-
tions of a particle m on the different points of a closed surface,
each resolved in the normal to the surface at the point.
Now, suppose the sm'face S to cut from a finite mass the
mass M^ the above equation holds for every element of this mass,
and therefore for the whole, if the symbols involved be properly
modified : we shall, therefore, still have the sum of the attrac-
tions on each point of ^S", resolved in the nonnal at that point,
= 47rJ/.
Again, suppose S to be an elementaiy parallelepiped so
small that the density [p] may be supposed uniform throughout
it: let V be the potential of a mass for an internal particle
whose coordinates are a?, y, z. Let P (fig. 121) be the point
a;, ^, s, and the comer of a parallelopipcd whose edges Zx^ Sy, S^,
arc parallel to the coordinate axes.
The above considerations shew that the sum of the attrac-
tions on the faces, each resolved in a direction pei-pendicular to
the face, will be due to the matter contained in the parallele-
piped: now
dV rrr p[x - ^) d^ dy d^
-III
dx- JJj{[j--^f+{^-r,Y+{z-^y]i^
368 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [18-48.
the integration extending throughout the parallelepiped ; hence
dV .
at P, -J— will be positive, since ^ is always greater than x ; at
P it will be negative, since ^ is always less than x + 8x.
Hence the absolute magnitude of the attractions parallel to the
axis of a^ at P and j) will be, respectively,
dV ^ dV d'V^
-J— and J =-T- ex.
ax ax ax
Now we may consider the surfaces il/P, mj) so small, that
the attraction on every point of each of them is the same : hence
the whole attractions on JWP, mp parallel to the axis of x
d^V
= -—hxhj 8z.
The whole expression for JJg cos Odo) is in this way found
to be
/ d'V d'V d'V\ . . ^
[-d^-df-^)^''^^ ^^' ^^"^^ ••• = ^'^^^
= 4:7rp 8x By Bz ;
d'V d'V d'V
•'- d^-^df-^d^ = - ^^^-
3. Supposing a mass of homogeneous fluid, which attracts
every particle of matter with a force varying as -.. ..;, to be
enclosed within a thin spherical shell, find the path described by
a heavy body let fall from any point of the surface of the fluid,
the resistance varying as the velocity. Prove also that the body
will reach the axis and equator of the spheroid after the same
intei-vals respectively, from whatever points of the surface it
begins to fall.
The attractions of the spheroid on any particle within it
perpendicular to the axis and equator, vary respectively as the
distances {x, y) of the particle from that line and plane, = ixx^ fix
suppose.
1848.] ATTKACTIONS. 369
Also, by the question, the resistance ou the particle in motion
= kv = k — : hence the resolved parts of it are
J ds dx - ds dti , dx , dii
dt ds ' dt ds ^ dt^ dt
Hence the equations of motion are
d X dx
^ + k^ + ;.x = 0,
d'?/ , dy
Let a, yS be the roots of the equation
z^ + kz + fi = 0,
and a', /9' those of
z"" + kz + yu,' = 0 :
the above equations give
X = Ae"" + Bt^',
y = A's*' + B'zl^'\
A, B, A\ B\ being arbitrary constants to be determined by the
circumstance that the body falls from rest from a given position.
The circumstance that it falls from rest gives us the condition
that ^ = 0, ^ = 0, when t=0:
dt ^ dt ^ '
.'. 0 = AoL + ^/3,
and 0 = A' a + B'^' ;
A B ^
.-. - = - - = C suppose,
A' B' „,
W ^~V^ suppose ;
the equations for the determination of the relation between
X and y by the elimination of t.
Hence, if f, t' be times of fallmg to the axis and equator
respectively,
0 = fit" - ai^\
BB
370 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1849.
.-. log/3 -\- at = loga + yS^;
log a — logyS
t =
so « =
a-y8 '
loga' — logyS' _
x' - ^' •
whence it appears that f, t' arc independent of the particle's
original position.
1849.
1. Each particle of two indefinite straight lines, lying in the
same plane, attracts with a force which varies inversely as the
distance. Determine the motion of a body projected in any
direction along the plane.
We must first find the attraction of either of the lines AB
(fig. 122) upon the particle in any position P. From P draw
PD, the perpendicular on AB^ and join PQ^ Q being a point
at the distance x from D. Let hA be the attraction of an
element hx of the line about Q resolved in PD:
.: BA^^cobQPB
fjbadx
.'. A = u tan~^ -
a
fJiTT.
from a; = — 00
to a; = + CO
Hence P will be attracted by two constant attractions in
constant directions, which are therefore equivalent to a constant
attraction in a constant direction, viz. 2yu,7r sin a (2a the angle
between the lines), parallel to the internal bisector of the lines.
Hence the case is the common case of projectiles, and the path
will be parabolic.
2. The attraction of a uniform filament of matter, in the
form of a plane curve, upon a particle is replaced by that of
a circular filament having the particle for its centre: find the
1850.] ATTRACTIONS. 371
law of density of the circular filament in order that this may
be done.
Let the cuitc be referred to the particle as pole, and let a
be the radius of the circle, yu. the density of the filament In the
form of the cun^e, p that of the circular filament at the point [B] ;
pa 80 fJ'^s ^
■' ^^ ,:^ >
fia ds
r
= fia \n^ +
[dej 1 '
u
1
7-
?
fia
^J'
if p be the
perpendicular from the
! particle
on
the
tangent
at the
point (r.
,0)
5
1
1850.
A uniform rod is placed witli its middle point against a rough
circle, in whose centre resides a force attracting inversely as the
square of the distance : if the rod be slightly disturbed from the
position of equilibrium, find the time of a small oscillation.
Let 6 be the inclination of the rod to the horizon at the
time t] ^ the distance of its middle point from the point of
contact with the circle. Since the motion is small, we may take
the equations of motion about the instantaneous axis of rotation :
hence we have
where k is the radius of gyration of the rod about its middle
point, and L the moment of the attractions on the rod about
the point of contact.
bb2
372 SOLUTIONS OF SENATE-HOUSE I'KOBLEMS. [I8f)l.
To find L. In fig. 120 the moment about D of the attrac-
tion on an element at Q (P being the centre of the circle),
if /i be the absolute force of the attraction, p the mass of a unit
of length of the rod,
fipa X Sx
{a' + xj'
_ , ^M- , 2? being the length of the rod,
{(i'^rf \^ ' a^ + r V d'^rj
omitting p and higher powers of |^,
2fipal g _ 2/jbpdH „
to the same degree of approximation : and the equation of
motion becomes
"^^ {d'+ry-
or, since M= 2?p,
g+ ^^ 0 = 0.
'^^ id'-^rfu'
Hence the time of a small oscillation
r=2
/Lt^a
1851.
1. Two uniform straight rods AB^ CD (fig. 123), mutually-
attracting each other with forces varying as the distance, are
1851.] ATTRACTIONS. 373
constrained to move in two grooves ABO, CDO at right angles
to each other ; dctenuine the time at which the extremity of one
of the rods reaches 0 the point of intersection of the grooves.
The attraction towards 0 of each element p'hri of CD upon
any element ph^ of AB, at a distance f from 0, will be the
same, viz. fip'SrjpB^.^ : hence, if AB = 2a, CB = 2b, the whole
attraction towards 0 of CB on AB will
r
= 2fipp'b
J -,
^dl.
if a? be the distance from 0 of the middle point of AB^
= Afipp'hax.
The equation of motion of AB is therefore
d'^x
2pa -y^= - ifipp'baxj
or -YY + w"'a: = 0, if n^ = 2fip'b ;
.'. x = A cos[nt + B)
= x^ cosntj
if t = 0 at the beginning of motion, and x^ is the original
value of X.
Hence, if t be the interval before A arrives at 0, we have
a = Xq cos nt,
1 -I a
or t = ,- — rfTi cos — .
{2fip'b)i x^
2. If a portion of a thin spherical shell, whose projections
upon the three coordinate planes through the centre are A, J5, C\
attract a particle at the centre with a force varying as any
function of the distance, shew that the particle will begin to
move in the direction of a straight line whose equatioas are
X y z
A^B^C'
Let 0 be the angle which the radius drawn to the element
SS of the shell makes with the axis of .r; then, if r = radius
374 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
of sphere, and </>(/•) be the law of attraction, the attraction
of 8S on the particle parallel to the axis of x will be
<j){r) 8S cosO^
and the whole attraction on it (X) parallel to the axis of Xj
X = <f){r) S.8/S'cos0, since r is constant,
= <f>{r)A.
So Y=<}>{r)B,
Z=<f>{r)C.
And the equations to the direction of the resultant attraction,
which is the direction in which the particle will begin to move,
are
X y z
X^ Y^Z'
X y z
( 375 )
PHYSICAL OPTICAS.
1848.
A spherical wave of light is incident directly on a lens : find
approximately the retardation of the several portions of the
• ,. , .111
wave, and prove m this way the common equation ~ / *
Suppose the lens to be a positive concavo-convex whose
thickness at the middle point is indefinitely small : take this
middle point as origin of coordinates and the axis of the lens
for axis of x. The retardation of any part or ray of the wave
will = (/u. — 1) X length of the path in glass = (/a — 1) p suppose.
Let ic, y be the coordinates of the point of incidence of this
ray, Q the inclination to the axis of the part of the ray within
the lens.
The equation to the two surfaces will be
V' = ^r^ (1),
and 77' = 2s| (2),
very nearly ; since | is very small for all rays near the axis.
Now (2) is satisfied by the coordinates x — p cos 9^ y- p sin 0,
or, as B is veiy small as well as p^ hj x — p and y ;
.-. f = 2s[x-p) ='-y-2sp',
1 ny
p = \r--sn
therefore the retardation of this portion of the wave
= (-')(7.-^)f-
Hence, if w be the distance from the origin of the centre
of the incident wave, the equivalent length in air of the ray
37(> SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1848.
we are considering from the centre of the wave to the point
of emergence,
= {(«-^)'+/]- + /*P
= u — X ■\- ^ +At \ — very nearly
2m \r h) 2 "^
Consequently this ray upon emergence is in the same phase as
the ray incident directly when it has travelled a distance
\ V ii\ V— alter emergence.
\u r ^\r s)\ 2 ^
Now the geometrical focus upon emergence is the centre of
curvature at the vertex of the surface of revolution, which is the
locus of all parts of the wave which, after transmission, are in
the same phase of vibration.
Let V be the radius of this sphere when only the parts of
the wave indefinitely near the axis have emerged : the sphere
will then pass through the points
[x-p, y) or (^1^, 3/j and
y being indefinitely small ;
u r ^\r syi 2 '
.-. / = 2v
11 ^1 _ l\\ t 4. t
u r^ *^\r s/j 2 "*" 25
11/ -N /^l 1
/. - = --f 0^-1
V u \r s
111 1/11
or = -^ , if -, ^ /A - 1)
V u J J \r s
But evidently v is the distance from the lens of the geome-
trical focus upon emergence ; hence this is the usual formvila.
1850.] PHYSICAL OPTICS. 377
1850.
A and B being two fixed points, and P such that AF= /m.BP,
the locus of P is a circle. Shew from this property how to
construct a lens of common glass, such that a direct pencil
incident from a determinate point will be refracted without
aben'ation.
The property enunciated will be found In Prob. 8, p. 157.
Let A, B (fig. 124) be the points from which the pencil is
to diverge before and after passing through the lens without
aberration. Draw the circular arc HCH' such, that If Q be
any point In It, BQ = fi.AQ.
With centre A describe any circular arc HcH' intersecting
HCH' in H^ H': HCH'c Is the section by the plane of the
paper of such a lens as Is required. For a ray incident upon
the lens from A will sufiier no deviation ; and
BQ-{fi.QP+AP) =fj,.AQ-{fi.QP+AP)
= {fi- l)AP Is constant:
and therefore, by reasoning similar to that in Airy's Tracts^
p. 276, it appears that the pencil diverging from A will, after
emergence, diverge from B.
1851.
If [6) be the angle which one of the planes of polarization
makes with the plane passing through the normal to the front of
the \f ave and either optic axis of a blaxal crystal, and y,, v^ be
the two velocities of transmission of the wave, shew that
(y,cos^)'''+ {v^%meY = h\
Since the planes of polarization respectively bisect the acute
and obtuse angles between the two planes through the normal to
the front and the optic axes (Griffin's Double Refraction^ Art. 21,
p. 12), It follows that the angle between these two planes = 2$.
Now, In accordance with the usual notation, the equations to
normal to the plane front are
? = J^ = ? (1).
I m n
378 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
Those to the optic axes are
3^ = ^^ (;^^4±(J^ = 0 (2).
If therefore tlie equations to the above planes are
Ax + Bi/ + Cz = 0,
A'x + i?'?/ + C'z = 0,
we must have
Al + Bm + Cn = 0,
A{d'-b-y + C{U'-c')^- = 0;
A B C _
Similarly
A ^b; C _ ,
AA' + BB' + CC
Also cos2^ =
{A + B' + cj {A" + B" + cy-
AA' + BB'+CC
~ {{AA'+BB'+CC'y+[BC'-CB'y+{CA'-AC'y+{AB'-BA'YY- '
Now
AA'+BB' + GC'= [m' {¥- c') + r {W- c^yd' {a'- b')-m' {d'-b')] rr'
= {[r-^m'){b'-c') - {m' + n'){d'-¥)]rr'
= {{r+m') b'-{l-n')6'- (I- P) d'+ {m'+n') &'} rr'
'.' P + m' + n' = 1
= _ {a' -b' + & - [I'd' + m'V' + d'c^) ] rr'
= - ( U— 25^) rr suppose,
[BC'-CB'Y + {CA'-AC'f + [AB'-BA'Y
= 4 {id'r' 4 m' + m'd') {d' - h')[h' - c') rV
= Anf{d'-b'){b'-c')rV'
= 4 Vr\"' suppose ;
1851.]
PHYSICAL OPTICS.
co-e '^' - ^
^«--^-{(2i^_?7r + 4F}*
2V - U
379
= (1 _ ^i^ _ ,„'-^) 7/c' + .ti^c'a' + {l-r- m'] d'h'
= W suppose ;
Now Vj"'*, vj'' are the roots of the equation
considered as a quadratic in v^ ;
.-. v^Jrv'^ = f [h' + c'O + m'^ (c^ + «•-=) + n'' {d' + J'^)
= (l_,,i-''_n'-')(J'''+c2)+(l_^--^_f-')(c''+a^)+(l_Z--'-7;i ){d'+l'')
= a'^ + ?>■■' + c' - {rd' + m^V' + ?i^c''')
and y>; = T^&^c'' + y«'''c'^«''' + d'd'W
. . cos2c/ =
v;' (1 + cos2^) + v^' (1 - cos 2^) = 2l/\
and (y^ cos^)' + ((',, sin 6)' = 6"^
( 380 )
CALCULUS OF YAIUATIONS.
1848.
A aud B are two given points in the generating line of
a surface of revolution whose axis is vertical : supposing a body
acted on by gravity, to descend along the surface from A to Bj
find its form when the whole pressure upon it between the two
given points is the least possible.
Find also the form of the surface when the length of the
generating line between the point A and B is also given, and
point out the difference between the two results.
Let y be the depth of any point of AB below A, x its dis-
tance from the axis ; the pressure at this point will be
P^Mg'^ + M-,
as p
V being the velocity, and p the radius of curvature at the point :
hence we must have
[Pds = Mg I f 1 + — . -^ I dx^ a minimum.
TT Tr , 2?/ ds
Here F = 1 + — -^
p dx
_i _ J£^.
dV _ 1q
i^ 1
dy 1 +^
dp (1+/)'^'
^ dq 1 +2^
1848.] CALCULUS OF VARIATIONS. 381
the equation between iV, P, Q^ is
^^ dP d'Q ^
or N--Up-^4)=0.
ax \ ax)
dQ _ 2p ^py
But ^-^ = -,-f^^ +
dx 1+/ ' (!+/)•''
or <? = 0,
shewing that the Une required is the straight line joinmg the
points A^ B.
K the length of the line be given, we have
a being an arbitrary constant to be determined.
N and Q remain of the same value as before ; P becomes
Hence we shall find, as before, that <? = 0, or the required
curve is a straight line : in this case, however, it must be a
broken line, the different parts of which are equally inclined to
the vertical, and the inclination so chosen as to give the line of
the required length. The particle is of course supposed to turn
the abiTipt angles of the line without impulsive pressure or
change of velocity.
382 SOLUTIONS OF SENATE-HOUSE PROBLEMS. [1851.
1851.
A uniform straight rod AB is, constrained to move in a
vertical plane with its middle point in a horizontal groove, and
its upper extremity against a smooth curve : find the nature of
the curve when the rod descends from one given position to
another in the least time possible, the initial angular velocity
being given.
Let CB (fig. 125) be the required curve, OQ the horizontal
groove ; take the point 0 in it for origin of coordinates : at time t
let AB be the position of the rod, draw BT the tangent at P;
OX = X, XB = 2/, 0Q = ^^ lOQB= e, QB = a.
The equation of vis viva is
-j-j + ^ \~ji) = constant = c suppose (1).
Also the motion of B pei'pendicular to BT'is zero,
.-. ^ miBTN +a~coB QBT = 0,
dt dt ^ '
or, if BTN=<j>, ^sin0 + a'^^cos(<^-6') = O (2);
also {^ + xY + / = d' (3).
From (1) and (2)
\ '^d'co%\<^-e)] \dt
or, since sin^ = -, tanrf> = ^ = w,
a dx ^^
X-W
Also, from (3)
" dx^ ' {d'-Tf-y
1851.]
CALCIJLUS OF VAKIATIONS.
383
dt
1 +
^•y
[[d^-yy+pyY\
d^
1
c[a-y
^a{(«^-j/?+P3/F+^4>?^^^.
or
Here V contains only y and p^
.-. V-Pp = C,
[{(« -y npy] +f^p^- [{{a^-yy+pyY^jc^^i + - (« -3^ ) -o,
{a^-yy{{^-yy+py] + 9. ^.f)i [[(^a^^,jy-+pyr+kYY- = o ;
1-
or {C'k-y)p = {d^-y% C = |l - (^)y';
.-. C'k sm" '^ + [p" - ff = x^ 0",
the required equation to the curve : the constants C", C" are to
be detemilned by the two given positions of the rod which give
two points through Avhich the curve must pass. The curve is
independent of the angular velocity of projection.
( 384
APPENDIX.
The following problem in Astronomy was set in 1848.
If a rectangular court be enclosed within a wall of given
height, and one of its sides be inclined at an angle of 30° to the
meridian, detennine the breadths of the shadows of the walls on
a given day at noon, and the portions of the courts and walls
which will be enveloped in the shadow, the latitude being
52° 30' north, and the Sun's declination on the given day
7° 30' north.
By referring to the problem on p. 64, we see that here
0 = 30° and </> = latitude — Sun's declination = 45°,
r. a = \h^ ^ " 2" ^''
Let ?j, ?2 ^6 the lengths of the walls, whose shadows are
respectively of the breadth «, 5, the area of the courts enveloped
in shade will be \a + [l^ — a) Z>, or l^a + I J) — ah ; and the
shaded parts of the walls the whole of the two walls, and two
triangles \ha^ \hh of the other two.
The following solution of the problem on p, 148, is due to
Mr. Gaskin.
Let TP, TQ (fig. 126) be the two given tangents, take the
line AB as axis of cc, and let OP' Q be the chord of contact of
any conic touching TP, TQ^ and passing through A^ B. Take
APPENDIX, 386
0 as origin, and let
'7 — ) ''7' —
a 0 a 0
be the equations to TP, TQ respectively ; also let OA = a,
OB = /3. Let the equation to OP' Q' he 7/ = mx, then that to
the conic will be
, Hence, putting ?/ = 0, we get
o,. 1, /i + i)l + (_L,_'|!) = o (.),
a; V« aj X \aa \J ^ '^
the roots of which equation are - , p 5 whence we see that
1111
a p a a
or the line OPQ is divided harmonically in -4, i?, whence 0 is
one of the foci of involution of the system of points P, Q^ A^ P,
so that the chords of contact of all conies touching PP, TQ and
passing tlu'ough -4, P, cut AB in one of the points 0, 0', if 0'
be the other focus of involution.
Now, in order that (1) may represent a rectangular hyperbola,
the sum of the coefficients of x'' and y'' must = 0 ; hence
-— - m^ + 777 - 1=0.
an 00
But by (2),
1 _ 1 m'
a/3 aa \
Combining these equations, we get
aa 00 J \aa ap
giving two values for «i, equal and of opposite signs^ so that
there can be constructed two pair of rectangular hyperbolae
CC
386 SOLUTIONS OF SENATE-HOUSE PROBLEMS.
whose chords of contact meet in one of the foci of invohition,
and are equally inclined to AB.
The relation between the four lines Ox, Oy, Op, Oq, in the
problem on p. 158, may be expressed thus: Ox and Oy each
bisect the lines between 0}), Oq parallel to the other.
For let the equations to Op, Oq, refen-ed to Ox, Oy as axes,
be 2/ = mx, y = m'x. Then if Oa = a, Ob = b, Od = a', Ob' = b',
the equations to ah, db' are
- +1 =1
a^b '
X y
a b
'.+'Tr=l;
whence, if (xy) be the point Q of intersection of these lines,
\a a J
But Q lies on the line Oq, or y = mx,
^b b'^
■'' («-^'Ka"a')='^^^"^'K^-T')'
11 /I lA ^
a a \b bj
The condition that the point of intersection of this line lies
on Op or y = mx, is derived from this equation by interchanging
a, a!, and writing m for m,
1 1 ,
? + m
a a
■(M)--
Hence m = — m, which expresses the above relation between
Ox, Oy, Op, Oq.
The same thing may be proved geometrically by making
any one of the points a, b, a, or b', remove to an infinite
distance.
APPENDIX. 387
The following statical Problems set in 1850 have been
omitted.
1. A heavy rod, whose weight is TV, rests upon a fulcrum
at its middle point, when loaded at one end with a weight W,
the density at any point of the rod at the distance x from a
TTtJC
certain point in it varies as sin — , a being the length of the
rod : find the ratio of W to TF', and determine at which point
the density is zero when this ratio is the greatest possible.
Let c be the distance from the centre of the rod of the point
where the density is zero, p the density at the point x = ^a.
The conditions of the problem give
r^ . TTic , r*-"-' . irx ,
p sm — ax ■\- [ p sm — dx = W,
^ Jo ^
pa (^ (tt 7rc\ , /tt ttcX) „_
or^|l-eos(- + -) + l-cos(j--)| = P^';
w W
••• P'^-^ (•)•
Also taking moments about the fulcrum which is at the
middle point of the rod,
r^"'' . . irx . r . , . TTcc
Pi (ic — c) sm — ax = p I [c- x) sm —
Jc ti '^J^^'a
TTX ,
ax
+ p \ (c + ic) sm — ax -\- W -
/•*'*' . irx J ( [*-"'' . irx J ^ fi"-' . irx J \ ^.., a
or pi xsm — ax — pel I sm — ax + I sm — ax ] = W -
T-T- f . TTX , ax TTX a^ . TTX
JN ow ixsm — ax = cos 1 — 5 sm \- c:
J a TT a TT a ^
. TTX , a . ire (a (a
icsm — rta; = -sm — \-z-\-c-\-\- -c
a IT « 2 V2
a . TTC
= — sm —
TT a
888 SOLUTIONS OF SENATE-HOUSE PROBLEMS.
and I sin — ax + l sin — ax = — , as shewn in (1
a'' . TTC 2 oca ,,,, a
r /IN TT W fa' . TTC 2ac\ „, a
or from (1 - — - sin \ = W -;
' 2 a Vtt a tt J 2
a sin 2c
a
This ratio = oo when c = 0 ; it has a maximum value when
. TTC .
a sm 2c is a mimmum ;
a
or, clIfFerentlating with respect to c,
TTC ^ ^
TT COS 2 = 0,
a
ire
COS — =
a TT
which determines the value of c.
2. Portions are cut from an ellipsoid by planes which are
parallel and equidistant from the centre ; if -ct be the length
of a pei-pendicular from the centre upon either plane, and
Z, m, ?i, the cosines of the angles which it makes with the axes,
shew that the remainder will rest when placed with a section
on a horizontal plane, if
1 r m' n'
— = or > -^ + -p- + ^ ,
w a o c
a, &, c, being the axes of the ellipsoid; and express the con-
dition that ]j such solids, when placed on each other with their
sections coincident, and their centres in a line inclined to the
vertical, shall not fall over.
The one portion will rest with a section upon a horizontal
plane if the vertical line drawn through its centre of gravity,
APPENDIX. 389
which is the centre of the ellipsoid, fall within the section;
i.e. if zj be equal or less than the radius vector (r) of the
ellipsoid drawn in the same direction, or since
if -^ = or > -^ = or > -, + 77 + -^ .
«j T a b c
Wlien there are ^; such solids placed on each other as above
described, the height of the centre of gravity above the plane
will be jjicr ; and, if p be the distance of the foot of the per-
pendicular drawn from the centre of gravity on the horizontal
plane from the centre of the section, p the same distance when
there is but one solid, we have p = i^p : the condition that the
p solids shall not fall over is, that pp shall be equal to or
less than the radius vector of the section through the foot
of the said perpendicular.
The equation to the cutting plane is
Ix + my + nz = vj (l) ;
and if a, /8, 7, be the coordinates of the centre of the section,
a, /8, 7, are subject to the conditions (see Gregory's Solid
Geometry^ Art. 121)
la + myS -f W7 = ■z^,
a /9 7 -ra- , s
and
ciH U'm c'n a'l' + H'm' + c^i'
The coordinates of the foot of the perpendicular on (1) are
IzTj v/jCT, w-ct; hence the equations to the radius vector of the
section through this foot are
y - ^ _ g - 7 _
= p,r suppose (3),
Izs — o. wa — ^ 7*^ — 7
where r is the distance of the point {xyz) fi'om (a/37).
If we substitute these values of ic, y, 2, in the equation ta
the ellipsoid
x' f _^ g' _ 1
i? + 6^ + c^-*'
390 SOLUTIONS OF SENATE-HOUSE PROBLEMS.
we shall find the length of the radius vector r of the section
through the foot of the perpendicular,
{{1^ - a) fir -\- aY {(yn^ - /3) fir + /3Y _^ {(»^ - 7) Z^^ + tF _ 1
7? ^ h' + 6' '~^'
the roots of this equation are equal ;
Now, from equations (2),
la 1)1/3 7i<y
a' "^ F "^ 6' aT + bW + 6'd' '
and -5 + 7T + -ii = -^7F
and from equations (3),
— = (?zj - a)' -I- [mzj - /Sy + {nzj - 7)"'
= w"'' - 2^' + a'"' + ^' + 7'
= d' + fi' + i' - ^'
Hence equation (4) becomes
{d' "^ U' "^ c"^ ^ dT + yW + d'd'] p' ~ dT + hW -f c^;i^ •
Now the equation of equilibrium is
p'
1
^
or
<
r
P'
henee the required condition is
= or < —, idH'' + Z^'^y/i'' 4 c'n^ - to'').
or |y (^1 + ^^ + ~) [d^l" + h\i^ -f 6^d^) - (/ - i)| to'
= or < dP + Z/^wi'^ + c'/i"'.
APPENDIX. 391
3. If a plane area, bounded by a parabola and its double
ordinate, be supported by an axis through the focus and a
vertical force acting along the ordinate, find what portion may
be cut oif by a line through the focus without aifccting the
vertical force ; and the least area for which this is possible.
Let P8Q (fig. 127) be the line cutting off the portion PQR :
the centre of gravity of PQR must lie in the vertical through 8^
or if we draw R V the diameter of PQ^ and SG vertical meeting
it in O^ G must be the centre of gravity of PQR. Hence
GV=IRV. Let AS=l, LPSB=e',
Also
But
or
tan^ =
21
SG'
GV =
SG cot^ =
2l coi'6
.
SP =
21
1 - cos^'
. PV =
SP- SV =
-. SP-
SG cosec
0
( 1
cos^X
sin'''<9J
21
Vl-cos^
21
^xxi'd'
PV =
sni d
4Z
~ sin' 6'
.5lcofe:
AP
Al.5lcoed\
!
.: cos''^ = i,
which determines the position of the line P8Q.
If the bounding ordinate have its extremity C nearer the
vertex than the point P just determined, let Q'SP' be the po-
sition of the cutting line, the centre of gravity of Q' CP' must
lie in the vertical through S: draw Q'u vertical: let AB' = a]
.'. I {2P-{x + l)i-xtand}xdx
= p^^{2Z4 {I -x)i + x tan e]xdx + 2(^ 2 (Ix)^ x dx.
J " Usectf
392 SOLUTIONS OF SENATE-HOUSE PROBLEMS.
This equation, when reduced, will detennine the value of 6
in teniis of a.
The least area for which this is possible will evidently be
sucli, that the part cut off will be the half, and the cutting line
A SB: in this case S is the centre of gravity of the whole area,
and AB = |Z.
The first part of the Prob. 5, on p. 212, may be proved by
referring to the values of the angles contained between any
two adjacent sides of a regular polyhedron (see Hall's Sjjherical
Trifjonometry^ Art. 59) : it appears that this angle is a sub-
multiple of 27r only in the case of the cube.
THE END.
CAMBRIDGE:
PRINTED BY METCALFE AND PALMER.
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